Prove the function is strictly increasing or decreasing Given that $f:(a,b) \rightarrow R$ is differentiable and $f'(x)$ is not $0$ for all $x$ in $(a,b)$.
I need to show that f is strictly increasing or decreasing.It seems trivial to me so I think I am missing something.
Since $f'$ is not $0$, it either greater than or less than 0. Then by the theorem $f$ is monotone increasing or decreasing.
Then it is proved.
Am I right? or did I miss something?
improvement
I think what I mentioned above does not prove strictness. Can I show strictness by saying $f$ is $g - h$ and $f'$ is $(g -h)'$ then since $f'$ is not $0$, $g$ is not $h$. So we can drop equality.
I now see where I am being dumb.
 A: Without a theorem, one could think that $f'(x)$ could jump from positive to negative (or the reverse) in the interval $(a,b)$ without ever becoming zero. After all, it is true that we do not know that $f'(x)$ is continuous, so the Intermediate Value Theorem for continuous functions does not apply.
However, we do have an Intermediate Value Theorem for derivatives, also called Darboux's theorem. Refer to that theorem and your problem becomes easy.3
ADDED:
Let $c=\frac{a+b}2$ and $a<x<b$. If $f'(x)$ has the opposite sign of $f'(c)$ then by Darboux's theorem $f'(d)=0$ for some $d$ between $c$ and $x$. This contradicts the hypothesis, and since $f'(x)$ cannot be zero, $f'(x)$ has the same sign as $f'(c)$. This is for any $x$, so $f'(x)$ is either all positive or all negative. Without loss of generality, let's say $f'(x)>0$ for all $x \in (a,b)$.
Take $x<y$ in $(a,b)$. By the Mean Value Theorem,
$$f'(e)=\frac{f(y)-f(x)}{y-x}$$
for some $e$ such that $x<e<y$. By hypotheses, both the left hand side and the denominator on the right hand side are positive. Therefore, $f(y)-f(x)$ is positive, thus
$$f(x)<f(y)$$
This was for any $x$ and $y$ such that $a<x<y<b$. This is the very definition of "$f$ is monotonic increasing in $(a,b)$". Similarly, if $f'(x)<0$ for all $x \in (a,b)$ then $f$ is monotonic decreasing in $(a,b)$. Either way, $f$ is monotonic.
There is no need for "tak[ing] a smaller interval."
A: Hint:


*

*Check out the Darboux's theorem.


I hope this helps $\ddot\smile$
