# Local Maximum of a Differentiable Function Using the Mean Value Theorem

• Benny
In summary, the conversation discusses a question about proving a local maximum for a function using the Mean Value Theorem. The participants discuss the necessary conditions for this proof and come to the conclusion that continuity at x_0 is sufficient, rather than differentiability as stated in the original question. The conversation also touches on the pronunciation of mathematical symbols.
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?

Benny said:
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.

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

Benny said:
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?

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?

I'm not sure but if f is not differentiable at x = x_0 but continuous 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.

WHAAT IS x_0 [phonetically] /?

extreme_machinations said:
WHAAT IS x_0 [phonetically] /?

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 $$x_f$$.

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.

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.

## What is MVT?

MVT stands for "Multivariate Testing". It is a statistical method used to test and compare multiple variables at the same time in order to determine the most effective combination.

## How can MVT help struggling students?

MVT can help struggling students by allowing educators to test and compare different teaching methods, materials, and interventions in order to determine the most effective approach for improving student performance.

## What factors are typically tested in MVT for struggling students?

Factors that are typically tested in MVT for struggling students include different teaching methods, study strategies, homework assignments, and learning materials (such as textbooks or online resources).

## How is MVT different from A/B testing?

MVT differs from A/B testing in that it allows for the testing of multiple variables at the same time, while A/B testing only compares two variables. MVT also takes into account the interactions between variables, while A/B testing focuses on individual variables.

## Are there any limitations to using MVT for struggling students?

While MVT can be a useful tool for improving student performance, it is important to keep in mind that there are many factors that can influence student success, and MVT may not always provide definitive answers. It is also important to consider ethical concerns and ensure that MVT is being used in an ethical and responsible manner.

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