Continuity and differentiability over a closed interval

[budafeet57]

Homework Statement

http://i.imgur.com/69BmR.jpg

Homework Equations

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?

 

[clamtrox]

b, d are right because f'(c) is differentiable over the interval

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

I am not sure about e. Can anyone explain to me?

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?

 

[HallsofIvy]

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

 

[budafeet57]

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?

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].

 

[budafeet57]

f(1)=4

 

[clamtrox]

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].

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.

 

[budafeet57]

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.

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?

 

[clamtrox]

I personally think drawing the picture is more informative

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.

Then D is the most interesting one: you should really think hard about this.