# 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]?

See explanation below.

By signing up, you agree to our Terms of Service and Privacy Policy

To use the Intermediate Value TheoremTo use the Intermediate Value Theorem toTo use the Intermediate Value Theorem to showTo use the Intermediate Value Theorem to show thatTo use the Intermediate Value Theorem to show that theTo use the Intermediate Value Theorem to show that the polynomialTo use the Intermediate Value Theorem to show that the polynomial function (To use the Intermediate Value Theorem to show that the polynomial function ( fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 +To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function (fTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \To use the Intermediate Value Theorem to show that the polynomial function (f(xTo use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cosTo use the Intermediate Value Theorem to show that the polynomial function (f(x)To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos xTo use the Intermediate Value Theorem to show that the polynomial function (f(x) =To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x )To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) hasTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 +To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has aTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = -1 + 3 \cos x ) has a root inTo use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3To use the Intermediate Value Theorem to show that the polynomial function ( 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 \To use the Intermediate Value Theorem to show that the polynomial function ( 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 \cosTo use the Intermediate Value Theorem to show that the polynomial function ( 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 xTo use the Intermediate Value Theorem to show that the polynomial function ( 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)To use the Intermediate Value Theorem to show that the polynomial function ( 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) hasTo use the Intermediate Value Theorem to show that the polynomial function ( 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 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, \To use the Intermediate Value Theorem to show that the polynomial function (f(x) = -1 + 3 \cos x) has a rootTo use the Intermediate Value Theorem to show that the polynomial function ( 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 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{To use the Intermediate Value Theorem to show that the polynomial function (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{5To use the Intermediate Value Theorem to show that the polynomial function (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\To use the Intermediate Value Theorem to show that the polynomial function (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\piTo use the Intermediate Value Theorem to show that the polynomial function (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}{To use the Intermediate Value Theorem to show that the polynomial function (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}{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,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}]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, \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}] \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, \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}] ),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{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 needTo use the Intermediate Value Theorem to show that the polynomial function (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 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\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 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\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 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}{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 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}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 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}]\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 withinTo use the Intermediate Value Theorem to show that the polynomial function (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 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}]), 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 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 needTo use the Intermediate Value Theorem to show 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 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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).

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.

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}] ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7