# Homework Help: Finding a Maximum

1. Nov 15, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Let f: U -> R where U is an open subset of R. Suppose that for some c in U, f'(c) = 0 and f''(c) < 0. Proof that the restriction of f to some open ball of center c attains a maximum at c.

The attempt at a solution
Let e > 0. Then for some d' > 0, if |x - c| < d' then |f(x) - f(c)| ≤ e|x - c| since f'(c) = 0. Let f''(c) = -b for some positive b. Then for some d'' > 0, if |x - c| < d'' then |f'(x) + b(x - c)| ≤ e|x - c|. Thus, for d = min{d', d''}, if |x - c| < d then

|f(x) - f(c)| ≤ e|x - c| and |f'(x) + b(x - c)| ≤ e|x - c|.

Of course, what I need is f(x) ≤ f(c). I don't see how to get that from the above. Any tips?

2. Nov 15, 2008

### Office_Shredder

Staff Emeritus
Assuming you know f'(x) > 0 implies f is increasing in a ball around x:

Sketch argument:
f''(c)<0 means f' is decreasing in a ball around c. Hence as f'(c)=0, f'(x)>0 if x<c and f'(x)<0 if x>c for x close to c. If c isn't a maximum, there exists x s.t. f(x)>f(c). -> in between x and c, there exists d s.t. f'(d) = f(x)-f(c)/(x-c) What can you say about the sign of f'(d)?

3. Nov 15, 2008

### e(ho0n3

I was not aware of this fact. I will attempt to prove it on my own.

If f'(d) is negative, x < c, but then we get a contradiction since f'(x) > 0 if x < c. A similar contradiction occurs if f'(d) is positive. So then f'(d) = 0 and f(x) = f(c), another contradiction. Thus c is a maximum.

Cool. Thanks a lot.

4. Nov 15, 2008

### e(ho0n3

This is more difficult than I thought. All I've been able to derive is the following: Let e > 0. Then there is a d > 0 such that if |y - x| < d, then

|f(y)| < (e + f'(x))|y - x| + |f(x)|.

But this doesn't help in determining that f(a) ≤ f(b) if a < b where a, b are in the ball of radius d centered at x.

5. Nov 15, 2008

### Office_Shredder

Staff Emeritus
If f'(y)>0 for some y, then $$\lim_{x \rightarrow y} \frac{f(x)-f(y)}{x-y} > 0$$

So heuristically (you can make this rigorous), near y, if x<y, f(x)<f(y) as otherwise the $$\frac{f(x)-f(y)}{x-y} < 0$$ and if x>y, f(x)>f(y) as otherwise $$\frac{f(x)-f(y)}{x-y} < 0$$ just by noting the numerator and denominator have opposite signs

6. Nov 15, 2008

### e(ho0n3

It never occured to me to look at [f(x) - f(y)] / (x - y). I understand your idea. However, that only shows that for some ball B centered at y, if x in B and x < y, then f(x) < f(y). It doesn't prove that f is increasing in B. For example, let x' in B and suppose x < x' < y. What can we conclude about f(x) versus f(x')?

7. Nov 16, 2008

### e(ho0n3

I don't think we need for f to be increasing. It's enough to show that f'(x)>0 if x<c and f'(x)<0 if x>c for x close to c.

Another comment: You seem to be using the mean value theorem for the existence of d, but we can't use it for we don't know whether f is continuous and differentiable on [x,c] (or [c,x]).