The function $f(x)$ exists in an open interval $(a,b)$ and is differentiable there. Let $c$ be a point that exists in the open interval $(a,b)$, and $f(x)$ has its maximum value at $x=c$. How do I prove that $f'(c)=0$?
(The unit name is mean value theorem.)
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Hint: Perturb $f(c)$ a bit to see that $f(c + h) \leq f(c)$ for some $|h| < \delta$ where $\delta$ is very small. Now what can you deduce when $h$ is positive and negative? Since $f$ is differentiable at $c$ what can you now conclude?
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If $f$ has a local maximum or minimum value at an interior point $c$ of its domain, and if $f$′ is defined at $c$, then $f'(c) = 0$.
To prove that $f'(c)$ is zero at a local extremum, we show first that $f'(c)$ cannot be positive and second that $f'(c)$ cannot be negative. The only number that is neither positive nor negative is zero, so that is what $f'(c)$ must be.
To begin, suppose that $f$ has a local maximum value (the proof for minimum is similar) at $x = c$ so that $f(x) - f(c) \leq 0$ for all values of $x$ near enough to $c$. Since $c$ is an interior point of $f$’s domain, $f'(c)$ is defined by the two-sided limit
$ \lim_{x \to c} \frac{f(x)-f(c)}{x-c} $
This means that the right-hand and left-hand limits both exist at $x=c$ and equal $f'(x)$. When we examine the limits separately, we find that
$f'(c)= \lim_{x \to c^{+}} \frac{f(x)-f(c)}{x-c} \leq 0$
$f'(c)= \lim_{x \to c^{-}} \frac{f(x)-f(c)}{x-c} \ge 0$
Together imply $f'(c)=0$.
Since $(a,b)$ is open, there exists $\sigma>0$ such that $c+h\in(a,b)$ for all $\sigma$ such that $|h|<\sigma$. Moreover, $$\Delta y=f(c+h)-f(c)\leq 0$$ Therefore, $\frac{\Delta y}{h}\geq 0$ if $-\sigma<h<0$ and $\frac{\Delta y}{h}\leq 0$ if $0<h<\sigma$.
Taking limit for $h\to 0^{-}$ e for $h\to 0^{+}$, we may conclude that $(D_{-}f(x))_{x=c}\geq 0$ and that $(D_{+}f(x))_{x=c}\geq 0$. Since $f$ is differentiable at $x=c$, we get that $$f'(c)=(D_{-}f(x))_{x=c}=(D_{+}f(x))_{x=c}=0$$
Same conclusion if $c$ is a minimum value point.