The required conditions for Intermediate Value Theorem include the function must be continuous and \(\displaystyle f(a)\)cannot equal \(\displaystyle f(b)\). While there is a root at \(\displaystyle (0,0)\) for this particular continuous function, this cannot be shown using Intermediate Value Theorem. The function does not cross the \(\displaystyle x\) axis, thus eliminating that particular answer choice. The correct answer is “No, because \(\displaystyle f(-1)=f(1)\).” Since one of the conditions for Intermediate Value Theorem is that \(\displaystyle f(a)\)cannot equal \(\displaystyle f(b)\), by graphing \(\displaystyle f(x)=x^2\) we can see that this requirement is not met.

his function is continuous (as it is a polynomial) and \(\displaystyle f(1)\neq f(3)\); therefore, the required conditions for Intermediate Value Theorem are met. While there is a root at \(\displaystyle x=2\) for this function (as can be seen by graphing the polynomial), Intermediate Value Theorem does not state where this root will be exactly, nor does it state how many roots there might be. Thus, the conclusion that can be made by IVT is that there is a root for this polynomial located somewhere between \(\displaystyle x=1\) and \(\displaystyle x=3\).

First, determine the values of the function at the bounds. This will allow the correct implementation of the Intermediate Value Theorem.

\(\displaystyle f(x)=x^2\)

\(\displaystyle f(2)=2^2=4\)

\(\displaystyle f(3)=3^2=9\)

Because the problem asks to analyze the interval \(\displaystyle [2,3]\) and \(\displaystyle f(2)=4< f(3)=9\), there must be a value \(\displaystyle f(a)< f(c)< f(b)\), with \(\displaystyle a< c< b\). Because \(\displaystyle f(2)=4< f(c)=7< f(3)=9\), by Intermediate Value Theorem there should be a number \(\displaystyle c\) between \(\displaystyle 2\) and \(\displaystyle 3\) that satisfies the required conditions. Therefore, “Yes, as shown by Intermediate Value Theorem” is the correct  answer.

If \(\displaystyle b=0\), only one root can be guaranteed (at \(\displaystyle x=0\)). 

If \(\displaystyle b= -1\), then Intermediate Value Theorem can be applied twice, for \(\displaystyle [-1,0]\) and \(\displaystyle [0,1]\).

This is true because for continuous functions, Intermediate Value Theorem states that a change in sign (ex: from positive to negative) of the function within an interval suggests a root (where the function crosses the \(\displaystyle x\) axis) at some point within that interval.

Intermediate Value Theorem states that if the function \(\displaystyle f\) is continuous and has a domain containing the interval \(\displaystyle [a,b]\), then at some number \(\displaystyle c\) within the interval \(\displaystyle [a,b]\) the function will take on a value \(\displaystyle f(c)=N\) that is between the values of \(\displaystyle f(a)\) and \(\displaystyle f(b)\).

The conditions that must be satisfied in order to use Intermediate Value Theorem include that the function must be continuous and the number \(\displaystyle c\) must be within the interval \(\displaystyle [a,b]\). However, \(\displaystyle f(a)\) cannot equal \(\displaystyle f(b)\).

Therefore, the answer choice “Intermediate Value Theorem can be used when \(\displaystyle f(a) = f(b)\)” does not satisfy the necessary conditions and is the correct answer for this question.

Intermediate Value Theorem is only true with continuous, differentiable functions, thus eliminating the answer choices “The function is not continuous within that interval” and “The function is not differentiable within that interval.” There is not necessarily a local maximum or minimum contained in the interval either. That leaves the correct answer choice, “There is a zero between the endpoints of that interval.” If the polynomial is changing signs and meets the requirements for Intermediate Value Theorem, it must cross the \(\displaystyle x\) axis at some point within the interval.

A polynomial has a zero or root when it crosses the \(\displaystyle x\) axis. For a given interval \(\displaystyle [a,b]\), if a and b have different signs (for instance, if \(\displaystyle a\) is negative and \(\displaystyle b\) is positive), then by Intermediate Value Theorem there must be a value of zero between \(\displaystyle f(a)\) and \(\displaystyle f(b)\). Therefore, Intermediate Value Theorem is the correct answer.

The Intermediate Value Theorem tells us that a value between \(\displaystyle f(a)\) and \(\displaystyle f(b)\) exists, but it does not provide any information on what that value is. This eliminates two of the four answer choices  - “The Intermediate Value Theorem can identify the value of \(\displaystyle f(c)=N\) that the function takes on as it passes from \(\displaystyle f(a)\) to \(\displaystyle f(b)\)” and “The Intermediate Value Theorem tells you how many times the function \(\displaystyle f\) repeats a value as it progresses from  \(\displaystyle f(a)\) to \(\displaystyle f(b)\).”

The Intermediate Value Theorem is useful because it can help identify when there are roots or zeros; an example of this is if a polynomial switches signs, Intermediate Value Theorem tells us there is a zero between those values.

From this, we can conclude that the correct answer is “The Intermediate Value Theorem states that somewhere between \(\displaystyle f(a)\) and \(\displaystyle f(b)\) there exists a value \(\displaystyle f(c)=N\), with \(\displaystyle a< c< b\).”

Graphing \(\displaystyle f(x)=x^3-5\) on cartesian coordinates reveals that the function is continuous and crosses the \(\displaystyle x\) axis at a value within the interval \(\displaystyle [1,2]\). 

Furthermore, setting \(\displaystyle x=1\) produces a negative value for the function, while setting \(\displaystyle x=2\) produces a positive value, as seen below:

\(\displaystyle f(x)=x^3-5\)

\(\displaystyle f(1)=(1)^3-5=1-5= -4\)

\(\displaystyle f(2)=(2)^3-5=8-5=3\)

Because the function is a polynomial, the function is continuous. By Intermediate Value Theorem, if the function changes signs within this interval, there must be a root present within the interval.

Because the function is a polynomial, the function is continuous. By Intermediate Value Theorem, if the function changes signs within this interval, there must be a root present within the interval. 

To apply Intermediate Value Theorem to the function \(\displaystyle f(x)=(x-1)^2-4\), the function can be evaluated at each of the given bounds. 

For instance, if the function is evaluated at \(\displaystyle x=-2\) and \(\displaystyle x=0\), the following is obtained:

 \(\displaystyle f(x)=(x-1)^2-4\)

\(\displaystyle f(-2)=(-2-1)^2-4=9-4=5\)

\(\displaystyle f(0)=(0-1)^2-4=1-4= -3\)

For the intervals \(\displaystyle [-1.5,-0.5]\), \(\displaystyle [-2,0]\), and \(\displaystyle [2,4]\), there is a change in sign within the interval.

Because the function does not change in sign within the interval \(\displaystyle [4,5]\), we cannot conclude by Intermediate Value Theorem whether there is a root contained in the interval or not. Thus, \(\displaystyle [4,5]\) is the correct answer.