**Q1: **

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 .

**Q2: **

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.

**Q3: **

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

**Q4: **

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.