I am confused about the following: $$\sqrt{x^2}=?\ (\sqrt x)^2 =?\ x^{2/2}$$ The source of confusion is: Let $$f(x)=\sqrt {x^2}$$ $$g(x)=\ (\sqrt x)^2 $$ Then $$f(-3)=3$$ $$g(-3)=\mbox{undefined}?$$
While the range of $f(x)$ and $g(x)$ might be the same, their domain appears to be different ($g(x)$ restricts $x$ to be positive). At the same time, both can be written as $$x^{2/2}$$
What am I missing? -Thank you :)
The problem is
$$\sqrt{x^2}=(\sqrt{x})^2$$
only if $x \ge 0$. If we are talking about real numbers.
In general we can write:
$$\sqrt{x^2}=\left(\sqrt{|x|}\right)^2 \Leftrightarrow \sqrt{x^2}=|x|$$
You are not missing anything. In general, the expression $x^a$ for $a \neq 0$ is usually defined only when $x \geq 0$ and on this domain, you have the exponential rule $x^{ab} = (x^a)^b = (x^b)^a$.
Sometimes you can extend the definition of $x^a$ so that it will be defined for all $x \in \mathbb{R}$. This happens for example if $a \in \mathbb{Z} \setminus \{ 0 \}$ or if $a = \frac{1}{n}$ when $n \in \mathbb{N}$ is odd but it doesn't happen if $a = \frac{1}{2}$. In any case, even if you can extend the definition, you noticed that the exponential rule stops holding in general for all $x \in \mathbb{R}$.
Thus, $x = (x^2)^{\frac{1}{2}} = (x^{\frac{1}{2}})^2$ for all $x \geq 0$ but it doesn't work for $x < 0$.
