Show $0 < f'(x) \leqslant \frac{1}{2}$

Question

Given $f(x) = \arctan(e^x)$, then $\displaystyle f'(x) = \frac{e^x}{1+e^{2x}}$. Clearly, $f'(x) > 0$ for all $x\in\mathbb{R}$.

Further, I want to show that $f'(x) \leqslant \frac{1}{2}$ for all $x\in\mathbb{R}$. How might I do this without appealing to a graph of $f'$?

Answer 1

You could find the $x$ for which $f''(x)=0$, which happens to be $x=0$, so $f'(x)\leq\frac{1}{2}$. Of course also check that it is a max either by plugging in a value on the left and on the right of $x=0$ or checking out $f'''$ (last one not recommended).

Answer 2

$$ \frac{e^x}{1+e^{2x}} \leq \frac{1}{2} \Leftrightarrow 2e^x \leq 1+(e^x)^2 \Leftrightarrow 0 \leq (1-e^x)^2$$

P.S. By the same trick you can prove that for all $a \in \mathbb R$ you have $$-\frac{1}{2} \leq \frac{a}{1+a^2} \leq \frac{1}{2}$$

Answer 3

In order to find the maximum value of $$\displaystyle f'(x) = \frac{e^x}{1+e^{2x}}$$ You may take the second derivative and solve for $$f''(x) =0$$

You will that the only critical point of $ f'(x)$ occurs at $x=0$ which is a local maximum with $f'(0)=1/2,$ which is also an absolute maximum due to its symmetry.

Thus you have $$ f'(x) \leq \frac{1}{2}$$ for all $x$.