# differentiable function with only positive/negative slope impies strictly increasing/decreasing

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

Since $$\exists x_3 \in (a,b) s.t. \forall x \in (a,b),0

We can conclude that

$$\exists x_3 \in (a,b) s.t. \forall x_1,x_2 \in (a,b),x_1

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:)

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

## 1 Answer

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 $$$$f'(c) = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$$$$ I'll leave it to you to write and finish up the proof properly.

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