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In my textbook they give the proof that a positive derivative implies an increasing function. Then they state this lemma (the converse)(see picture) but give no proof. I was wondering if anyone could guide me in the right direction. All the proofs I have tried just seem like a copy of the original proof that the positive derivative implies an increasing function. However, I don't believe that is a really satisfying answer - seems rather recursive.

This is what I've come to:

Let $ \epsilon = g'(t_0)$. Then there exists a neighborhood $N$ of $t_0$ s.t. for $t \in N$ and $ t \ne t_0$ we have that $|\frac {g(b)-g(a)}{(b-a)} -g'(t_0)| < g'(t_0)$ so $\frac {g(b)-g(a)}{(b-a)}$ is positive for $t \in N$, $ t \ne t_0$.

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  • $\begingroup$ If $g'(t)<0$ for some $t\in(a,b)$ you should be able to show $g$ is not increasing near $t$, by using the definition of differentiability of $g$ at $t$. $\endgroup$ – kimchi lover Sep 14 '17 at 2:37
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    $\begingroup$ Fix $t \in (a,b)$. For a small positive $h$, $g(t+h) \geq g(t)$ since $g$ is increasing. Can you rearrange this to say something about the derivative? $\endgroup$ – xk3 Sep 14 '17 at 2:40
  • $\begingroup$ I was thinking of using the definition of the derivative and showing that for $\epsilon > 0$ we have that $g' (t) > -\epsilon$. However, I am not entirely sure this would work. $\endgroup$ – conums Sep 14 '17 at 19:50
  • $\begingroup$ Let $c \in(a, b) $. Can you see that if $f$ is increasing then the ratio $(f(x) - f(c)) /(x-c) $ is non-negative? $\endgroup$ – Paramanand Singh Sep 17 '17 at 3:38
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If $h>0$, we have for $t\in (a,b)$,
$$ g(t+h)-g(t)\geq 0\implies\frac{g(t+h)-g(t)}{h}\geq 0 $$ So $$ \lim_{h\to 0^+}\frac{g(t+h)-g(t)}{h}\geq 0 $$ Similarly, for $h<0$ we have $$ g(t+h)-g(t)\leq 0\implies\frac{g(t+h)-g(t)}{h}\geq 0 $$ so $$ \lim_{h\to 0^-}\frac{g(t+h)-g(t)}{h}\geq 0 $$ and finally $$ \lim_{h\to 0}\frac{g(t+h)-g(t)}{h}=g'(t)\geq 0 $$

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