Show Function is increasing

Question

Suppose that $x > 0$. Given a function $f(x) = x(1 - \mathrm{e}^{-x})$, what is the easiest mathematical way to find out if this function is monotone?

It is easy to plug numbers on $x$ and see that it is increasing for $x > 0$. But how would you prove it? Should I take the derivative of $f(x)$? Or $\log(f(x))$? Please provide the proof in steps.

Answer 1

The easiest way is to compute $f'(x)$, which happens to be equal to $e^{-x}(e^x+x-1)$. But $x>0\implies e^x>1$. So, $x>0\implies f'(x)>0$ and therefore $f$ is monotonic.

Answer 2

Differentiating $f(x)=x(1-e^{-x})$ gives:

$$f'(x)=(1-e^{-x})-\left[-xe^{-x}\right]=1+e^{-x}(1+x)$$ Clearly $e^{-x}\ne \forall x\in\mathbb{R}$. Now for $x \gt0$, $f'(x)\gt0$ since $x+1$ is clearly increasing and $e^{-x}\ne0$ so you get that $f$ is monotonic on $\mathbb{R^+}$.