Let $f:I\rightarrow \mathbb{R}$ be a differentiable function where $I$ is an open interval. Let $a,b\in I$ be such that $f'(a)<0$ and $f'(b)>0$. Then there is $c$ between $a$ and $b$ such that $f'(c)=0$.
Proof: (1) We show that there is local minimum $c$ for $f$ in the interval $(a,b)$.
(2) If (1) is not true, then either $c=a$ will be a local minimum or $c=b$ will be a local minimum.
(3) Suppose $c=a$ is local minimum. Then $f(a+h)-f(a)\ge 0$ in small neighbourhood $[a,a+h)$ of $a$, hence $\lim_{h\rightarrow 0^+} \frac{f(a+h)-f(a))}{h}\ge 0$ i.e. $f'(a)\ge 0$, contradiction.
(4) If $c=b$ is a local minimum, then $f$ is decreasing in local neighbourhood of $b$ i.e. $f(b)\le f(b-h)$ for small neighbourhood $(b-h,b]$ of $b$. But then $\lim_{h\rightarrow 0^+} \frac{f(b-h)-f(b)}{-h}\ge 0$ i.e. $f'(b)\le 0$, contradiction.
(5) Thus, local minimum must be inside $(a,b)$ and consequently, $f'(c)=0$.
Q. In the whole argument, we tried to find local minimum. It is natural question to ask why don't we try for local maximum. If $c=a$ or $c=b$ is a local maximum, then the arguments as in (3) and (4) do not give any contradiction actually. So we can not conclude that local maximum does or doesn't exist in $(a,b)$.
On the other hand, we can give an example of a function, such as $f(x)=x^2$ for $x\in [-1,1]$, where $f'(-1)<0$ and $f'(1)>0$. The point $c$ of local maximum is the boundary point $1$, and hence it is not in $(-1,1)$.
Geometrically, how can we justify that we should seek for local minumum but not local maximum to get desired $c$?
Best Answer
This is because it's assumed that $f'(a)<0$ and $f'(b)>0$. If you assume that $f'(a)>0$ and $ f'(b)<0$, then you will have to look for a global maximum.