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In the textbook, Mathematical Analysis of Apostol, there is the intermediate value theorem for derivative as shown below,

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Now after the theorem there is the note that this theorem is also true if one or both of the one sided derivatives at the endpoints are infinite, and to prove that Apostol suggests using this function $ g(x) = f(x) - cx$. I understand the theorem and the proof of it but how do I use this function $g$ above to show the result??

I can show this a different way just by extending the result of the theorem above to cover for the infinite one sided derivatives but I still wonder how to use that function $g$ as suggested to show that?? any hints would be great. thank you.

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Let us explicitly assume that $f$ is continuous on $[a,b]$. Then $g(x) = f(x) -xc$ is continuous on $[a,b]$ and differentiable on $(a,b).$ $g$ is increasing at $a$ and decreasing at $b$. Therefore it must assume its maximum at some point $x^*$ in the open interval $(a,b).$ By Theorem 5.9 (page 109) of the same book we have $g'(x^*)= f'(x^*) -c =0$.

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    $\begingroup$ Thank you for the reply. I got your idea but I just want to ask, from the fact that we know $g$ is increasing at $a$ and decreasing at $b$, how can we conclude that $g$ assume its maximum at some point $x'$ in the open interval $(a,b)$? I mean intuitively it should be like that but how do I show that rigorously? thank you. $\endgroup$ – Khoa ta May 8 '16 at 19:34
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    $\begingroup$ Without loss of generality assume $f'(a) > f'(b)$. Then $g'(a) > 0$ and $g'(b) < 0.$ From the first inequality we see that $g$ is increasing at $a$ and therefore $a$ cannot be a point where $g$ attains its maximum value [for $x >a$ and $x$ close enough to $a$, $g'(a) > 0$ implies $g(x) > g(a)$], similarly $g'(b) < 0$ implies that for points $x$ very close to $b$ we must have $g(x) > g(b)$ and for this reason $g$ cannot attain its maximum value at $b$. This leaves the interior points $(a,b)$ where $g$ must attain its maximum value. $\endgroup$ – dsm May 8 '16 at 19:40
  • $\begingroup$ thank you for your response. But what I meant is that how do we conclude that a local max even exist in $(a,b)$, not whether the local max occur at endpoint or at the interior. It seem true to me since $g$ is increasing at $a$ and decreasing at $b$ that $g$ should have a local max somewhere in $(a,b)$ but how do I show that rigorously?? $\endgroup$ – Khoa ta May 8 '16 at 19:50
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    $\begingroup$ $[a,b]$ is compact and $g$ is continuous. Thus $g$ must attain a maximum value on this compact interval (see "preservation of compactness" for continuous functions). $\endgroup$ – dsm May 8 '16 at 19:52
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    $\begingroup$ it's clear to me now, I totally forgot about that theorem. thank you for your help. $\endgroup$ – Khoa ta May 8 '16 at 19:54

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