How do you use the Intermediate Value Theorem to show that the polynomial function # f(x) = (x^3)/2 - 4x + 1/(x)# has a root in the interval [-3, -1]?

Answer 1

See below

The Intermediate Value Theorem states that, if a continuous function over an interval #[a,b]# is such that #f(a)=x# and #f(b)=y#, then #f# assumes all possible values between #x# and #y# in the interval #[a,b]#. This function is mostly used to find zeroes of a function, by finding a point in which the function is negative, and one in which it's positive: passing from negative to positive (i.e., from below to above the #x#-axis), the function must cross it, and that's the root.
So, we must verify that #f(x)# is continuous in #[-3,-1]#, and that #f(-3)# and #f(-1)# have opposite signs.
Coefficients apart, the first two terms are power of #x#, and so they are continuous everywhere. The third term, #1/x#, has its only discontinuity in #x=0#, which is outside of our domain #[-3,-1]#, so #f# is indeed continuous in #[-3,-1]#.

Now we only need to compute:

#f(-3)=(-3)^3/2 - 4*(-3) + 1/-3 = -11/6<0#
#f(-1)= -1/2 -4(-1) -1 = 4-1-1/2 = 5/2>0#.
Sign up to view the whole answer

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

Sign up with email
Answer 2

To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = \frac{x^3}{2} - 4x + \frac{1}{x} ) has a root in the interval ([-3, -1]), we need to demonstrate that the function changes sign within that interval.

First, evaluate ( f(-3) ) and ( f(-1) ).

( f(-3) = \frac{(-3)^3}{2} - 4(-3) + \frac{1}{-3} = \frac{-27}{2} + 12 - \frac{1}{3} = -\frac{45}{6} + \frac{72}{6} - \frac{2}{6} = \frac{25}{6} > 0 )

( f(-1) = \frac{(-1)^3}{2} - 4(-1) + \frac{1}{-1} = -\frac{1}{2} + 4 - 1 = \frac{7}{2} > 0 )

Since ( f(-3) > 0 ) and ( f(-1) > 0 ), and ( f(x) ) is a continuous function, by the Intermediate Value Theorem, there exists at least one root of ( f(x) ) in the interval ([-3, -1]).

Sign up to view the whole answer

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

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7