1. May 4, 2005

Benny

Hello everyone, I'm stuck on a MVT question. Can someone please help me out? Its not really a homework question, I'm doing this question to enhance my understanding of various things.

Q. Where a < x_0 < b, suppose that f(x) is differentiable in (a,b) and f'(x_0) = 0. Suppose also that for some delta > 0 , f'(x) > 0 for all $$x \in \left( {x_0 - \delta ,x_0 } \right)$$ and f'(x) < 0 for all $$x \in \left( {x_0 ,x_0 + \delta } \right)$$. Prove that f(x) has a local maximum at x = x_0.

Here is what I have done. I have left out some minor steps so that people can easily and quickly understand my working.

Consider a function f(x) which is continuous on [a,b] and differentiable on (a,b) as given in the the question.

Let $$x_0 - \delta < x_0 + \delta$$ be in [a,b](btw for this particular question does it make any difference if I say let ... be in [a,b] or (a,b)?).

Case 1: By the MVT we have

$$\frac{{f\left( {x_0 + \delta } \right) - f\left( {x_0 } \right)}}{{\left( {x_0 + \delta } \right) - x_0 }} = f'(d),d \in \left( {x_0 ,x_0 + \delta } \right)$$

$$\Rightarrow f\left( {x_0 + \delta } \right) - f\left( {x_0 } \right) < 0$$ since f'(x) < 0 for $$x \in \left( {x_0 ,x_0 + \delta } \right)$$.

So $$f(x_0 ) > f\left( {x_0 + \delta } \right)...(1)$$.

In a similar manner(I left out the working associated with the following statement to make this post more concise), using the MVT on the interval $$\left( {x_0 - \delta ,x_0 } \right)$$ it can be shown that:

$$f(x_0 ) > f\left( {x_0 - \delta } \right)...(2)$$

From (1) and (2) it can be seen that f(x_0) > f(x) for all x 'near' x_0 and the given result follows? This is the part I am worried about. It seems to me it is sufficient to show that f(x_0) >= f(x) for x near x_0 but I have only been able to get to a strictly greater than condition. Moreover, I haven't used the given equation f'(x_0) = 0 so my answer is obviously incorrect. Can someone please help me produce a proper answer to this question?

2. May 4, 2005

shmoe

So you've shown something stronger, that's fine! If a>b then a>=b.

What if we changed the condition of f'(x_0) = 0 to say "f is not differentiable at x_0" but left everything else the same. Would the conclusion still be true? If so how would you prove it?

3. May 5, 2005

Benny

I'm not sure but if f is not differentiable at x = x_0 but continous at x = x_0, and everything else stayed the same, since I treated the cases of x < x_0 and x> x_0 seperately, using the MVT, my conclusion that f(x_0) > f(x) near x = x_0 should still hold true. I'm not sure how to prove it though.

4. May 6, 2005

extreme_machinations

WHAAT IS x_0 [phonetically] /?

5. May 9, 2005

motai

x_0, also denoted as $$x_0$$, is usually pronounced as "x not" and refers to the initial condition of a variable (in contrast to the final variable $$x_f$$.

6. May 9, 2005

mathwonk

your "obviously incorrect" answer is actually correct. i.e. the hypotheses were far too strong. you only needed continuity at x0, (and differentiability elsewhere).

I.e. applying MVT to the interval [xo-delta, xo] shows that f is increasing there, and applying it to the other side shows f is decreasing there. QED.

7. May 10, 2005

Benny

Thanks for the confirmation mathwonk. I should have thought about what I was actually doing instead of just reading too deeply into the wording of the question.