How do you use the Intermediate Value Theorem to show that the polynomial function #f(x) = -1 + 3 cos x# has a root in the interval [0, 5pi/2]?

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the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need toTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to showTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

FirstTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show thatTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First,To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that theTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, noteTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the functionTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note thatTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changesTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes signTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign onTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(xTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on thatTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that intervalTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

FirstTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) isTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note thatTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is aTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuousTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous functionTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function becauseTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) isTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 +To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuousTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhereTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cosTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

NextTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next,To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluateTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\leftTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\fracTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

NextTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, weTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we findTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\rightTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\leftTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\fracTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cosTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) =To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\rightTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 +To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\leftTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\fracTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdotTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), theTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) =To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the functionTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changesTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes signTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 +To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign onTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on theTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the intervalTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) =To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \fracTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

SinceTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]),To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) =To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which meansTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by theTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the IntermediateTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate ValueTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value TheTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) andTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, thereTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there existsTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\leftTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists atTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at leastTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\fracTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least oneTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (cTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) inTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) =To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \fracTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ),To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), andTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and theTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the functionTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( fTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}])To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(xTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) suchTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x)To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such thatTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) isTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(cTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuousTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0),To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) andTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), iTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e.,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \fracTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(xTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem,To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in theTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there existsTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the intervalTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists atTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval \To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at leastTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least oneTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one valueTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0,To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \fracTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( cTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) inTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\piTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) in the intervalTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) in the interval (To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) in the interval ( [0To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]\To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) in the interval ( [0,To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]), we need to show that the function changes sign on that interval.

First, note that (f(0) = -1 + 3 \cos(0) = -1 + 3 \cdot 1 = 2).

Next, we find (f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3 \cdot 0 = -1).

Since (f(0) = 2) and (f\left(\frac{5\pi}{2}\right) = -1), the function changes sign on the interval ([0, \frac{5\pi}{2}]), which means, by the Intermediate Value Theorem, there exists at least one (c) in ([0, \frac{5\pi}{2}]) such that (f(c) = 0), i.e., (f(x) = -1 + 3 \cos x) has a root in the interval ([0, \frac{5\pi}{2}]).To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root in the interval ( [0, \frac{5\pi}{2}] ), we need to demonstrate that the function changes sign within that interval.

First, note that ( f(x) ) is a continuous function because it is a polynomial and ( \cos x ) is continuous everywhere.

Next, evaluate ( f(0) ) and ( f\left(\frac{5\pi}{2}\right) ). ( f(0) = -1 + 3 \cos(0) = -1 + 3(1) = 2 ) ( f\left(\frac{5\pi}{2}\right) = -1 + 3 \cos\left(\frac{5\pi}{2}\right) = -1 + 3(0) = -1 )

Since ( f(0) = 2 ) and ( f\left(\frac{5\pi}{2}\right) = -1 ), and the function ( f(x) ) is continuous between ( 0 ) and ( \frac{5\pi}{2} ), by the Intermediate Value Theorem, there exists at least one value ( c ) in the interval ( [0, \frac{5\pi}{2}] ) such that ( f(c) = 0 ), which means ( f(x) ) has a root in the interval ( [0, \frac{5\pi}{2}] ).