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Benny

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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 [tex]x \in \left( {x_0 - \delta ,x_0 } \right)[/tex] and f'(x) < 0 for all [tex]x \in \left( {x_0 ,x_0 + \delta } \right)[/tex]. 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 [tex]x_0 - \delta < x_0 + \delta [/tex] 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

[tex]

\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)

[/tex]

[tex]

\Rightarrow f\left( {x_0 + \delta } \right) - f\left( {x_0 } \right) < 0

[/tex] since f'(x) < 0 for [tex]x \in \left( {x_0 ,x_0 + \delta } \right)[/tex].

So [tex]f(x_0 ) > f\left( {x_0 + \delta } \right)...(1)[/tex].

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 [tex]\left( {x_0 - \delta ,x_0 } \right)[/tex] it can be shown that:

[tex]f(x_0 ) > f\left( {x_0 - \delta } \right)...(2)[/tex]

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?