Let $X$ be a random variable with probability density function $$f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \text{ for each } x \in \mathbb{R}.$$ We want to find the probability density function of $Y := X^2$.
I already found the expression of $f_Y$ (correct me if I'm wrong, please):
We know that $$F_Y(y) = P[Y \leq y] = P\left[X^2 \leq y\right] = P\left[\left|X\right| \leq \sqrt{y}\right] = \\$$ $$P\left[-\sqrt{y} \leq X \leq \sqrt{y}\right] =^{(1)} P\left[-\sqrt{y} < X \leq \sqrt{y}\right] = F_X\left(\sqrt{y}\right) - F_X\left(-\sqrt{y}\right).$$ Hence $$f_Y(y) =^{(2)} F'_Y(y) = F'_X\left(\sqrt{y}\right)\frac{1}{\sqrt{y}} = f_X\left(\sqrt{y}\right)\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{2 \pi y}}e^\frac{-\left|y\right|}{2}.$$
But where is the domain of $f_Y$? I mean, obviously $y > 0$, so would $\mathbb{R}^+$ be the domain of the function? I think I could determine the support of $Y$ before finding the expression, but how?
Other secondary questions:
(1) This is because $P[X = -\sqrt{y}] = 0$, right?
(2) I think I can only say this if $f_Y$ is continuous. Why would I assume continuity?
I don't see very well the absolute value thing, because we have to \begin{align*} F_{Y}(y) & = P( Y \leq y) \\ & = P(X^2 \leq y) \\ & = P(|X| \leq \sqrt{y}) \\ & = P(- \sqrt{y} \leq X \leq \sqrt{y}) \\ & = F_{X}(\sqrt{y}) - F_{X}(-\sqrt{y}) \\ f_{Y}(y) & =\frac{1}{2\sqrt{y}} f_{X}(\sqrt{y}) + \frac{1}{2\sqrt{y}} f_{X}(-\sqrt{y}) \\ & = \frac{1}{2\sqrt{y} \sqrt{2\pi}} e^{- (\sqrt{y})^2 /2} + \frac{1}{2\sqrt{y} \sqrt{2\pi}} e^{- (-\sqrt{y})^2 /2} \\ & = \frac{1}{\sqrt{y} \sqrt{2\pi}} e^{- y /2} \end{align*} Now, for the support of $Y = X^2$ we can do two things. The first is to graph the transformation $y = x^2$, this is
like $x \in \mathbb{R}$, we focus on the $y$ axis, but it goes from $0$ to $\infty$, the $y > 0$. And another somewhat less formal is to separate the interval and evaluate it in the transformation. $$ A_1 = (-\infty, 0) , \quad A_2 = \{0\}, \quad A_3 = (0,\infty) $$
we evaluate
$$ \lim_{x \to - \infty} x^2 = \infty $$ $$ \lim_{x \to \infty} x^2 = \infty $$
$$ \lim_{x \to 0} x^2 = 0 $$ then $y > 0$. Well, if we square all the possible values of $x > 0$, they are still $x > 0$
Remember that in practice it doesn't matter if you touch $0$ or not, that can be easily fixed.