If $f:(a,b)\rightarrow\mathbf{R}$ is differentiable and
1.$f'(x)>0$ for all x $\in (a,b)$.
$\ \ $then $f$ is strictly increasing on (a,b).
2.Similarly, if $f'(x)<0$ for all x $\in (a,b)$, then $f$ is strictly decreasing on (a,b)
Proof.(1.)
Assume $f:(a,b)\rightarrow\mathbf{R}$ is differentiable
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ and $f'(x)>0$ for all x $\in (a,b)$
Show that $\forall x_1,x_2 \in (a,b),x_1<x_2\rightarrow f(x_1)<f(x_2)$
Since $\exists x_3 \in (a,b) s.t. \forall x \in (a,b),0<f'(x_3)\leq f'(x)$
We can conclude that
$\exists x_3 \in (a,b) s.t. \forall x_1,x_2 \in (a,b),x_1<x_2 \rightarrow f(x_1)<f(x_1)+f'(x_3)(x_2-x_1)\leq f(x_2)$
Therefore, $f$ is strictly increasing on $(a,b)$
Proof.(2.)
Similar to (1.)
Questions:
1.Is my proof right?
2.Any other ways to prove this?
Thanks:)
Here's a "hint". Pick any $x_1, x_2 \in (a,b)$ such that $x_1 < x_2$. By the mean value theorem, there's a $c \in (x_1, x_2)$ such that \begin{equation} f'(c) = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \end{equation} I'll leave it to you to write and finish up the proof properly.