Show that $\sinh(x)$ is strictly monotonic.

Question

Is it enough to show that its derivative, $\cosh$, has no zeroes? (No $x$ satisfies $e^x=-e^{-x}$)

Answer 1

You should show that the derivative is strictly positive.

You could solve it as

$$ \sinh(x) = \frac{e^x - e^{-x}}{2} \\ \frac{d \sinh(x)}{dx} = \frac{e^{x} - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} $$

Which is always greater than $0$, since $e^x > 0 \ \ \forall x \in \mathbb{R}$

Since the derivative exists all over $\mathbb{R}$, the function $sinh(x)$ is continuous as well, which tells you that it is monotone increasing.