Homework Help: Continuity and differentiability over a closed interval

1. Oct 18, 2012

budafeet57

1. The problem statement, all variables and given/known data
http://i.imgur.com/69BmR.jpg

2. Relevant equations

3. The attempt at a solution
a, c are right because f(c) is continuous.
b, d are right because f'(c) is differentiable over the interval
I am not sure about e. Can anyone explain to me?

Last edited by a moderator: Oct 18, 2012
2. Oct 18, 2012

clamtrox

Are you sure about that? What about a linear function? Then f'(x) = constant ≠ 0.

What the claim is saying is that there exists a well-defined maximum value for f. How would the function f need to behave for this claim not to be true?

3. Oct 18, 2012

HallsofIvy

You appear to have "cut off" a critical part: what is f(1)?

4. Oct 18, 2012

budafeet57

Ah I see, b is the answer. Because f is not a constan, so it can only be linear. When it's differentiated it can't equal to zero under [-2,1].

Last edited: Oct 18, 2012
5. Oct 18, 2012

budafeet57

f(1)=4

6. Oct 18, 2012

clamtrox

That makes no sense at all. Why won't you draw a picture to get some idea of what's happening, if you want to reason like this without any mathematical proofs.

7. Oct 18, 2012

budafeet57

Thank you clamtrox. I am very rusty about the definition of continuity and differentiability, even I went back to my calc textbook, I still cannot figure out what I should do. Can you tell me what's wrong with my reasoning?

8. Oct 19, 2012

clamtrox

I think the problem in your reasoning is that you don't fully understand the problem. So visually, these things you know about the function:
-it starts from p=(-2,-5) and ends in q=(1,4)
-there are no jumps
-there are no sharp edges
You are asked to check if there exists any such function for which the conditions A-E do not hold.

A is easy: you cannot draw a curve from p to q without crossing f=0. This can be shown easily using intermediate value theorem.

Now for B: here you need to check if all curves from p to q must have f'=0 at some point. In otherwords, they are parallel to the x-axis. Now I already gave you one counterexample for why this isn't true. I am sure you can draw several other curves that also are not parallel with x-axis at any points, but still satisfy the other conditions.