Worksheet: The Mean Value Theorem and Its Interpretation

The mean value theorem is one of the most important theoretical tools in calculus. It is used if the function is defined and continuous on a known interval and differentiable on that interval. This worksheet assesses your understanding of the mean value theorem and its uses.

Q1:

Madison is not convinced that the mean value theorem is true because, she says, the function is certainly differentiable on . But if we take and , we have , and yet there is no point where . What is her error?

AThe function should be strictly decreasing on the interval.
BThe theorem requires that the function be differentiable everywhere, which is not.
CThe function should be strictly increasing on the interval.
DThe theorem requires the domain to be an interval, which is not.
EThe function is not continuous, and the theorem requires continuity on an interval.

Q2:

Consider the result: If is differentiable on an interval and , then , a constant, for all .

Which of the following statements says exactly the same thing as the constant function result?

AIf is a constant function on an interval , then is differentiable and at all .
BIf is differentiable on an interval and not a constant function, then at all .
CIf is differentiable on an interval and at some , then is not a constant function.
DIf is differentiable on an interval and not a constant function, then at some .
EIf is differentiable on an interval and at each , then is not a constant function.

If is differentiable on an interval and not constant, we get points with . How does this show that at some point ?

Abecause then and by the mean value theorem, there is a point with
Bbecause if on , then would be a constant function
Cbecause only constant functions has everywhere
Dbecause means that one of or is not zero; we can take as this point
Ebecause then and by the mean value theorem, there is a point with

Q3:

Mason is not convinced that the mean value theorem is true because, he says, the function has the property that if we take and , we have , and yet there is no point where . What is his error?

AThe function should be strictly decreasing on the interval.
BThe theorem requires the domain to be an interval, which is not.
CThe function should be strictly increasing on the interval.
DThe function is not differentiable at . The theorem requires diferentiability on an interval.
EThe function is not continuous. The theorem requires continuity on an interval.

Q4:

Consider the statement that if is a differentiable function on an interval and there, then is strictly increasing on .

Which of the following statements is equivalent to the above?

AIf is differentiable on an interval and there, then is not strictly increasing on .
BIf is differentiable on an interval and not strictly increasing there, then at all points .
CIf is differentiable on an interval and strictly increasing there, then at all points .
DIf is differentiable on an interval and not strictly increasing there, then at some point .
EIf is differentiable on an interval and strictly increasing there, then at some point .

What does it mean for a function to not be strictly increasing on the interval ?

AThere are points with but .
BThere are points with but .
CThere are points with but .
DThere is a point where .
EWhenever satisfy , then .

Using the equivalent statement to the main result, how can you use the mean value theorem to prove the equivalent statement?

AIf is differentiable on and not strictly increasing, then take with . By the mean value theorem, we get between and where , so .
BIt is not possible to prove the statement using the mean value theorem.
CIf is differentiable on and not strictly increasing, then take with . By the mean value theorem, we get between and and where , so .
DIf is differentiable on and not strictly increasing, then take with . By the mean value theorem, we get between and where , so .
EIf is differentiable on and not strictly increasing, then take with . By the mean value theorem, we get between and where , so .