# MVT - please help struggling student

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.

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