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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:)

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    $\begingroup$ The statement $\exists x_3 \in (a,b) s.t. \forall x \in (a,b),0<f'(x_3)<f'(x)$ is totally unjustified (and in fact, wrong). $\endgroup$ – Eric Wofsey Jun 16 at 4:24
  • $\begingroup$ It is also totally unclear how you reached the conclusion that $\exists x_3 \in (a,b) s.t. \forall x_1,x_2 \in \mathbf{R},x_1<x_2 \rightarrow f(x_1)<f(x_1)+f'(x_3)(x_2-x_1)\leq f(x_2)$. $\endgroup$ – Eric Wofsey Jun 16 at 4:24
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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.

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  • $\begingroup$ Yes, that's so true. Otherwise, it would be contradiction. thanks $\endgroup$ – Manx Jun 16 at 4:44

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