# Intro to analysis proof first and second derivatives and mean value theorem

1. Feb 12, 2010

### reb659

1. The problem statement, all variables and given/known data

Let f(x) be a twice differentiable function on an interval I. Let f''(x)>0 for all x in I and let f'(c)=0 for some c in I. Prove f(x) is greater than or equal to f(c) for all x in I.

2. Relevant equations

Mean value theorem?

3. The attempt at a solution

f''(x)>0 implies that f'(x) is strictly increasing on I. I don't know what to use the f'(c)=0 for, and whether or not to use the mean value theorem.

2. Feb 12, 2010

### Kizaru

What does f''(x) indicate about f(x)? When you see this, you will realize how the f'(c)=0 comes in to play.

3. Feb 12, 2010

### reb659

It implies that f(x) is concave up?

It also implies c is a minimum by the second derivative test but we haven't covered that as of this section in our textbook.

4. Feb 13, 2010

### reb659

I still can't figure this one out - I get the feeling the proof is much more simple than I think it is.

5. Feb 13, 2010

### Dick

Apply the MVT. If f(c) is not a minimum then there is a d such that f(d)<f(c). Suppose d<c. What does the MVT tell you? You are right, the proof is not that hard.

6. Feb 13, 2010

### reb659

The MVT tells me that there exists some Xo in [d,c] such that f'(Xo) = [f(c)-f(d)]/(c-d).

7. Feb 13, 2010

### Dick

Sure. Is that quantity positive or negative?

8. Feb 13, 2010

### reb659

It is negative. I'm still a bit puzzled as to how the mean value theorem relates to proving this problem.

Last edited: Feb 13, 2010
9. Feb 13, 2010

### Dick

I don't agree with that. We assumed f(d)<f(c) and d<c. I think it's positive. You'll see how this problem relates to the MVT shortly. Feel free to jump ahead if you like.

Last edited: Feb 13, 2010
10. Feb 13, 2010

### reb659

Ah my mistake, the quantity is positive. I think I know where this is headed.

So f'(xo)>f'(c)=0 and c>xo. But f'(x) must be strictly increasing on I since f''(x)>0, and thus the statement must be true by contradiction.

11. Feb 13, 2010

### Dick

Yes, exactly. That's where the f'(c)=0 comes in. And if d>c?

12. Feb 13, 2010

### reb659

If d>c then f'(xo)<0 and xo>c. f'(c)=0 and f'(xo)<0. But this contradicts that f'(x) is strictly increasing. So the statement must be true.

13. Feb 13, 2010

### Dick

So you have it then.

14. Feb 13, 2010

### reb659

Thank you so much for your help!