Increasing function and its derivative.

Question

I know that:

1. $f(x)$ is increasing $\iff$ $f'(x) >= 0$
2. $f'(x) > 0$ $\to$ f(x) is strictly increasing (not the other way around)

But I have read in a true/false homework that:

If $f]0,1[ \to R$ is derivable and strictly increasing on $]0,1[$ then $\to$ $f'(x)$ is increasing on $]0,1[$ is a FALSE statement

I don't understand it because if the function is strictly increasing then we know that it is increasing also so $f'(x) >= 0\space ?$ so why is it false?

Answer 1

The idea of this exercise is to point out that while $f$ being increasing and $f'$ being positive are intricately linked, $f$ being increasing and $f'$ being increasing are not. For instance, see what happens if $f(x) = 2x-x^2$. $f$ is strictly increasing, and $f'(x) = 2-2x$ is strictly positive, which is as it should be. However, $f'$ is decreasing.

Answer 2

In the statement

Given: $f(x)$ is strictly increasing.

To Check: $f'(x)$ is also increasing.

Solution: $f(x)$ is strictly increasing only implies that $f'(x)\geq0$.

For proving $f'(x)$ to be increasing we have to see if $f''(x) \geq0$ but we don't know the value of $f''(x)$. Hence we can't say that $f'(x)$ is an increasing function.