If I have $$f(x) =\sqrt{g(x)}$$
Will the variations of $f(x)$ be the same than the variations of $g(x)$ ?
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The definition of variation has... well... variation. Can you explain what "variation" means? – Shahar Sep 18 '14 at 19:56
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What exactly do you mean by 'variations'? Changes in $f(x)$ and $g(x)$ will not be the same: $f(x)-g(x)\neq f(x)-\sqrt{g(x)}$. – lemon Sep 18 '14 at 19:56
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By "variations" I mean the way the values of the function goes up and down (for example the variations x^2 is decreasing on ]-∞;0[ and increasing on ]0;+∞[ – Pop Flamingo Sep 18 '14 at 19:57
2 Answers
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.
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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,
$$ \frac{d}{dx} \sqrt {g(x)} = \frac{g'(x)}{2 \sqrt {g(x)} }$$
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