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I tried to prove the theorem myself, that if for every $x$ in the domain $f'(x)$ is positive, then the function $f$ is strictly increasing. This is how it went:

$$f(x_1) < f(x_2)$$

$$f(x) < f(x+h), h>0 $$

$$f(x) < f(x) + f'(x)\cdot h + O(h), \lim_{h \rightarrow 0} \frac{O(h)}{h} = 0$$

$$0 < f'(x) \cdot h + O(h) $$ $$0 < f'(x) + \frac{O(h)}{h}$$

Now here is where I am not sure about the correctness of what I am about to do next...to take the limit of both sides and then:

$$ 0 < f'(x) + 0$$

Is the proof correct ?

VLC
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    This isn't a proof, it's a list of formulas with no explanation of how they are connected. Furthermore, you appear to be reasoning backwards. Don't start with what you're supposed to prove. Start with what you know, and reason from there to what you're supposed to prove. So start with "Assume that $f'(x) > 0$ for all $x$ in some interval $I$. Suppose $x_1, x_2 \in I$ and $x_1 < x_2$." Now explain (in sentences, not just a list of formulas) why it must be the case that $f(x_1) < f(x_2)$. – Dan Velleman Jul 03 '21 at 17:05
  • What about the Lagrange mean value theorem? – PinkyWay Jul 03 '21 at 17:09

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You actually attempted to prove the converse of the statement (counterexample given below). In order to prove the desired statement, you would start with $f'(x) > 0$ for all $x$ in the domain of $f$, and try to conclude $f(x_1) < f(x_2)$ for all $x_2 > x_1$.

Jeff Cheng
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