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Let $f:Irightarrow mathbbR$ be a differentiable function where $I$ is an open interval. Let $a,bin I$ 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 $lim_hrightarrow 0^+ fracf(a+h)-f(a))hge 0$ 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 $lim_hrightarrow 0^+ fracf(b-h)-f(b)-hge 0$ 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 $xin [-1,1]$, 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 $c$?

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.

It is best to understand the meaning of sign of derivative. Let's see what the assumptions $f'(a) <0,f'(b)>0$ mean. Since $f'(a) <0$ the function $f$ is strictly decreasing at point $a$ which more formally means that there is an $h>0$ such that $f(a) >f(x) $ for all $xin(a, a+h) $. The graph of $f$ thus looks to be going downwards at $a$ as we move to the right of $a$. Similarly since $f'(b) >0$, the graph of $f$ moves upward at $b$ as we move from the left of $b$. If you draw such a graph on paper you will at once be convinced that there is local minimum of $f$ between $a$ and $b$.

The conclusion can be reached without any graphical aid. Assume $f(a) leq f(b) $ (the case $f(a) >f(b) $ can be handled similarly). Since there are values of $f$ in interval $(a, a+h) $ which are less than $f(a) $ and also less than $f(b) $ because $f(a) leq f(b) $ it follows that the minimum value of $f$ in $[a, b] $ is attained at an interior point and this also acts as a local minimum.

The conditions in the question do not prohibit the existence of a local maximum and there may (or may not) be a local maximum also, but the conditions don't guarantee it. The local minimum on the other hand is guaranteed as shown in previous paragraph.

You are right, $f'(c)=0$ may imply either minimum or maximum. For example, $y=frac14x^4-frac12x^2, xin (-2,2)$.

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$f(0)=0$ is a local maximum. However, there are two global minimums, namely: $f(-1)=f(1)=-frac14$.

The proof relies on the Extreme Value Theorem and assumes that the function attains its maximum and minimum in the closed interval, therefore it checks the three points: the point $c$ and the borders $a$ and $b$. It proves that the borders cannot be minimum, therefore the minimum occurs inside at the critical point $c$ for which $f'(c)=0$. And note there can be several points inside the interval for which $f'(c)=0$, but surely there exists such $c$ for which the function attains its minimum.

A real point of confusion here is that you are assuming that a < b in your proof [witness the description of the open interval as '(a,b)', which notation assumes a < b]. The problem statement DOES NOT state this!

If a < b as you assume, the conditions imply that the graph of f descends from left-to-right from a, and ascends from left-to-right to b, which as you infer puts a local minimum between them. However, if b < a, the graph of f -ascends- from left-to-right from b and -descends- from left-to-right to a, implying a local -maximum- between them.