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If we have $f(x)=\sqrt{x}$

Find $g(x)$ where $f(g(x))=-f(x) =-\sqrt{x}$

I know that the square root can have plus and minus value in the same time

but I what that $f(g(x))$ value is negative when $f(x)$ is positive and vice versa.

My calculations led me to controversial results that $g(x) = (-1)^2x$

When I plug it back to my equation I found my self confused between two different ways to solve the problem.

The first solution

$f(g(x))=f((-1)^2x)=\sqrt {(-1)^2x}=(-1)^{2*\frac 1 2} x^{\frac 1 2}=-\sqrt{x}$

Or the second confusing solution

$f(g(x))=f((-1)^2x)=\sqrt {(-1)^2x}=\sqrt {1*x}=\sqrt{x}$

Actually I wonder if there is a role control this situation and show what the priority of handling powers with bases, I mean shall I handle powers with powers first or handle powers with base first, and if the second solution is correct then $g(x)$ must be other value to have a negative result, so for that I am really confused.

Pentapolis
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  • $f(x)=\sqrt{x}\ge0$, but $f(g(x))=-\sqrt{x}\le0$. There can be no solution except at $x=0$. – robjohn Aug 25 '16 at 22:32
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    "the square root can have plus and minus value in the same time" This is a common confusion. A function can only take a single value. By convention the square root function $\sqrt x$ denotes the positive square root. See these previous questions: http://math.stackexchange.com/q/26363/856, http://math.stackexchange.com/q/41878/856 –  Aug 25 '16 at 22:33

2 Answers2

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The only solution is $g(x)\equiv0$. Since $f(x)=\sqrt{x}$, we know $f(x)\geq0$ for all $x\geq0$. Furthermore, we know $f(g(x))=-\sqrt{x}\leq0$, which implies that $f(g(x))=\sqrt{g(x)}=0$. Solving this gives $g(x)=0$.

Clayton
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If you analyse the function $f$ you find that $f:\mathbb{R}_0^+ \rightarrow\mathbb{R}_0^+$. So you got that $f(g(x))=-\sqrt x=-f(x) \Leftrightarrow 0=f(g(x))+f(x)\ge0+0=0$. From there we get that $f(g(x))=f(x)=0$, which is only true for $x=0$.

CryoDrakon
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