All Calculus AB Resources
Example Questions
Example Question #1 : Apply Intermediate Value Theorem
Which of the following does NOT satisfy the conditions required to apply the Intermediate Value Theorem to a function on the interval ?
The function f must be continuous along to apply Intermediate Value Theorem
Intermediate Value Theorem considers points within the interval
Intermediate Value Theorem cannot be applied outside of the interval
Intermediate Value Theorem can be used when
Intermediate Value Theorem can be used when
Intermediate Value Theorem states that if the function is continuous and has a domain containing the interval , then at some number within the interval the function will take on a value that is between the values of and .
The conditions that must be satisfied in order to use Intermediate Value Theorem include that the function must be continuous and the number must be within the interval . However, cannot equal .
Therefore, the answer choice “Intermediate Value Theorem can be used when ” does not satisfy the necessary conditions and is the correct answer for this question.
Example Question #1 : Apply Intermediate Value Theorem
Using Intermediate Value Theorem to analyze a continuous function, what can be deduced if a polynomial changes signs within an interval?
The function is not differentiable within that interval
The function is not continuous within that interval
There is a local maximum between the endpoints of that interval
There is a zero between the endpoints of that interval
There is a zero between the endpoints of that interval
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 axis at some point within the interval.
Example Question #1 : Apply Intermediate Value Theorem
What theorem could you use to show that a polynomial has a root on a given interval?
Mean Value Theorem for Derivatives
Fundamental Theorem of Calculus
Extreme Value Theorem
Intermediate Value Theorem
Intermediate Value Theorem
A polynomial has a zero or root when it crosses the axis. For a given interval , if a and b have different signs (for instance, if is negative and is positive), then by Intermediate Value Theorem there must be a value of zero between and . Therefore, Intermediate Value Theorem is the correct answer.
Example Question #1 : Apply Intermediate Value Theorem
Using the continuous function and the interval , which of the following correctly identifies why the Intermediate Value Theorem is useful?
The Intermediate Value Theorem tells you how many times the function repeats a value as it progresses from to
The Intermediate Value Theorem states that somewhere between and there exists a value , with
The Intermediate Value Theorem can identify the value of that the function takes on as it passes from to
The Intermediate Value Theorem is not useful
The Intermediate Value Theorem states that somewhere between and there exists a value , with
The Intermediate Value Theorem tells us that a value between and 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 that the function takes on as it passes from to ” and “The Intermediate Value Theorem tells you how many times the function repeats a value as it progresses from to .”
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 and there exists a value , with .”
Example Question #1 : Apply Intermediate Value Theorem
In which interval does have a root?
Graphing on cartesian coordinates reveals that the function is continuous and crosses the axis at a value within the interval .
Furthermore, setting produces a negative value for the function, while setting produces a positive value, as seen below:
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.
Example Question #3 : Apply Intermediate Value Theorem
In which interval does the function NOT necessarily have a root?
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 , the function can be evaluated at each of the given bounds.
For instance, if the function is evaluated at and , the following is obtained:
For the intervals , , and , there is a change in sign within the interval.
Because the function does not change in sign within the interval , we cannot conclude by Intermediate Value Theorem whether there is a root contained in the interval or not. Thus, is the correct answer.
Example Question #1 : Apply Intermediate Value Theorem
Can Intermediate Value Theorem be applied to the function within the interval ?
Yes, because the function crosses the axis within the interval
No, because the function is not continuous
No, because
Yes, because the function has a root at
No, because
The required conditions for Intermediate Value Theorem include the function must be continuous and cannot equal . While there is a root at for this particular continuous function, this cannot be shown using Intermediate Value Theorem. The function does not cross the axis, thus eliminating that particular answer choice. The correct answer is “No, because .” Since one of the conditions for Intermediate Value Theorem is that cannot equal , by graphing we can see that this requirement is not met.
Example Question #3 : Apply Intermediate Value Theorem
What can be concluded by using Intermediate Value Theorem for the function on the interval ?
There are two roots on this polynomial located between and
The requirements for Intermediate Value Theorem are not met
There is a root for this polynomial located between and
There is a root for this polynomial at
There is a root for this polynomial located between and
his function is continuous (as it is a polynomial) and ; therefore, the required conditions for Intermediate Value Theorem are met. While there is a root at 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 and .
Example Question #2 : Apply Intermediate Value Theorem
Let . Is there a number between and such that ?
Yes, as shown by the Fundamental Theorem of Calculus
No, Intermediate Value Theorem cannot determine the exact value of
No, no number c such that exists
Yes, as shown by Intermediate Value Theorem
Yes, as shown by Intermediate Value Theorem
First, determine the values of the function at the bounds. This will allow the correct implementation of the Intermediate Value Theorem.
Because the problem asks to analyze the interval and , there must be a value , with . Because , by Intermediate Value Theorem there should be a number between and that satisfies the required conditions. Therefore, “Yes, as shown by Intermediate Value Theorem” is the correct answer.
Example Question #2 : Apply Intermediate Value Theorem
Assume is continuous on the interval and has the values listed in the table below. Which of the following values of guarantees that has at least two roots?
If , only one root can be guaranteed (at ).
If , then Intermediate Value Theorem can be applied twice, for and .
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 axis) at some point within that interval.