power and radical functions

Question

I have 2 functions: $f(x) = x^2$ and $s(x) = \sqrt{x}$. I need to determine which of the following 3 statements is/are true for all $x$: $$$$ a) $\quad f(s(x)) = x \qquad$ b)$\quad s(f(x)) = x \qquad$ c) $\quad (s(x))^2 = f(x)$

All/specific/none statements can be true. According to my calculations all of them are true but I'm not sure.

Answer 1

Hint:

remember that $\sqrt{x}$ is defined only for $x\ge 0$ and is always not negative, so : $$ \sqrt{x^2}=|x| $$ and $$ \left(\sqrt{x} \right)^2=x $$

Answer 2

Hint:

Set $x=-1$ and tell me which of the statements is true.

Answer 3

Observe that Dom$\,(f)=\Bbb R\;$ and Im$\,(f)=[0,\infty)\;$ , , whereas Dom$\,(g)=[0,\infty)=$ Im$\,(g)\;$ .

Now, for general functions $\;h,k\;$ , the composition $\;h\circ k\;$ is defined iff Im$\,(k)\subset$ Dom$\,(h)\;$ , so in our case both compositions are defined but only for positive values of $\;x\;$ , as we can see from the above, and then yes:

$$\begin{align*}&f\circ s(x):=f(s(x))=f(\sqrt x)=(\sqrt x)^2=x\\{}\\&s\circ f(x):=s(f(x))=s(x^2)=\sqrt{x^2}=x\end{align*}$$

since $\;\sqrt{x^2}=|x|\;$ and we're working in the above only with $\;x\ge 0\implies |x|=x\;$ .

For negative values of $\;x\;$ the composition $\;f\circ s\;$ isn't defined since $\;s(x)\;$ isn't defined, though it is defined

$$s\circ f(x):=s(f(x))=s(x^2)=\sqrt{x^2}=|x|=-x\;,\;\;\text{since}\;\;x<0$$

Now you answer your three questions there.