Square root of increasing function is increasing

Question

How can it be shown mathematically that the square root of an increasing function is also increasing? Logically, it makes sense since the square root function is an increasing function. Does it suffice to show that the first derivative is positive?

Answer 1

Note that if $f(x)$ is increasing, then $$x_1\le x_2 \implies$$

$$ f(x_1)\le f(x_2) \implies $$

$$\sqrt {f(x_1)}\le \sqrt {f(x_2)}$$

Thus if $f$ is increasing so is $\sqrt f $

Answer 2

It does suffice, but note that if your function is not differentiable then you can't do this.

For a more general answer, prove that the composition of increasing functions is increasing. (This is true whenever $f: X \to Y$, $g: Y \to Z$ are increasing, regardless of what the orders are on $X, Y, Z$.)

Let $f, g$ be increasing. Suppose $x < y$. Then $g(x) < g(y)$ since $g$ is increasing. So $fg(x) < fg(y)$ since $f$ is increasing.