Does the square root function change the variations of a function?

Question

If I have $$f(x) =\sqrt{g(x)}$$
Will the variations of $f(x)$ be the same than the variations of $g(x)$ ?

Answer 1

First, I'll assume that $g$ is a real valued function defined on some real interval, and that $g(x)\geq 0$ for all $x\in\text{Dom }g$.

Since the function $\sqrt{\cdot}$ is increasing, we have that $\sqrt{g(x)}\geq \sqrt{g(y)}$ if and only if $g(x)\geq g(y)$. So the answer is yes, $g$ and $f$ have the same "variations", or using a more standard language, they are increasing or decreasing in the same intervals. (By the way, "increasing" is the usual word for a function "going up").

Remark: Note that I haven't said that words variation or go up/down aren't correct. The problem is that they are not standard, that is, commonly accepted. In order to a better communication, it is better using standard words, or if not, defining the non-standard ones.

Answer 2

Not sure you know calculus; since $\sqrt x$ is always increasing, we find that $\sqrt {g(x)}$ is increasing wherever $g(x)$ is increasing, decreasing wherever $g(x)$ is increasing, and horizontal ( a "critical point") wherever $g(x)$ is horizontal. I suggest you draw some graphs on actual graph paper, you might call it quadrille paper...

Here, draw $$ y = x^4 + 4 x^3 + 28 $$ for $-4 < x < 3, $ then draw $$ y = \sqrt{x^4 + 4 x^3 + 28}. $$ These have horizontal inflection points when $x=0,$ minima when $x=-3.$

Meanwhile, with the hypothesis that $g(x) > 0$ always on our interval of definition,

$$ \frac{d}{dx} \sqrt {g(x)} = \frac{g'(x)}{2 \sqrt {g(x)} }$$