Prove that if $f'(c)>0$ (strictly positive) for all $c\in(a,b)$ then $f(x)$ is strictly increasing on $(a,b)$

Question

Prove that if $f'(c)>0$ (strictly positive) for all $c\in(a,b)$ then $f(x)$ is strictly increasing on $(a,b)$

It is true that if $f(x)$ is strictly increasing on $(a,b)$ then $f'(c)>0$ (strictly positive) for all $c\in(a,b)$? If it is true, prove it. If it is not, find a counter example.

Let $c \in (a,b)$ be arbitrary we will show $f'(c)>0$. Consider a sequence $x_n \rightarrow c$. Since $f$ is differentiable at $c$ we have that $\frac{f(x_n)-f(c)}{x_n-c} \rightarrow f'(c)$. Now since $f$ is strictly increasing and $x_n \rightarrow c$ we have that $f(x_n)>f(c)$.($f(x_n) \not = f(c)$ since $f$ is continuously getting larger). Thus $\frac{f(x_n)-f(c)}{x_n-c}>0$ and therefore $f'(c)>0$.

I feel the second is also true but not sure what to add.

Answer 1

The second part, that if $f(x)$ is strictly increasing on $(a, b)$ then $f'(c) > 0$ for all $c \in (a, b)$, is not necessarily true. Consider $f: (-1, 1) \to \mathbb R$ where $f(x) = x^3$ as a counterexample. Your proof is flawed in that you assume $$ (\forall n)\ \frac{f(x_n) - f(c)}{x_n - c} > 0 \implies f'(c) = \lim_{n \to \infty} \frac{f(x_n) - f(c)}{x_n - c} > 0$$ Limits do not preserve strict inequalities, just weak ones. Hence your proof does show that $f'(c) \geq 0$, which is true. But $f'(c) > 0$ is too strong a result.

A big hint to prove the first part, that if $f'(c) > 0$ for all $c \in (a, b)$ then $f$ is strictly increasing, is to use the mean value theorem.

Answer 2

Assume $f$ is differentiable on the open interval $(a,b)$ (so that $f'(c)$ makes sense for any $c\in (a,b)$) then $f$ is differentiable on any open subinterval $(x,y)\subseteq (a,b)$. By the mean value theorem there exists $c\in (x,y)$ such that $$\frac{f(y)-f(x)}{y-x}=f'(c)\Rightarrow f(y)-f(x)=f'(c)(y-x)>0\Rightarrow f(y)>f(x)$$ since by assumption $f'(c)>0$ (and also $y>x$). Therefore $f$ is strictly increasing on the interval $(a,b)$ since $(x,y)$ was arbitrary.

Also for the second question: if $f$ is strictly increasing then $f(y)>f(x)$ whenever $y>x$. Then by definition of the derivative $$f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\geq 0.$$

Edit: Counterexample to strict inequality as provided by @bitesizebo.