Q: Let $X$ be a continous random variable s.t. $X\sim U(-1,3)$. Find $F_Y$, for $Y=X^4$. Q: Let $X$ be a continous random variable s.t. $X\sim U(-1,3)$. Find $F_Y$, for $Y=X^4$.
My attempts so far:
$F_Y(t)=P(Y\le t)=P(X^4\le t)=P(-\sqrt[4]{t}\le X\le\sqrt[4]{t})=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.
Since $X\sim U(-1,3)$
then $F_X(t)=\begin{cases}0&,  t<-1 \\ \frac{t+1}{4}&,  -1\le t\le 3 \\ 1&, 3<t &\end{cases}$
What I don't understand is how find the right way to split $F_Y(t)=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.

Let's step back for a minute and think about what happens to $Y$ for various values of $X$.
When $-1 \le X \le 1$, then $0 \le Y \le 1$.  So when we pick some $y$ between $0$ and $1$, $$\Pr[Y \le y] = \Pr[-y^{1/4} \le X \le y^{1/4}].$$  But when $1 < y \le 3^4$, instead of writing $-y^{1/4} \le X \le y^{1/4}$, the left-hand side inequality stops at $-1$ because that's where the support of $X$ ends.  And when $y > 81$, then both left and right sides of the inequality becomes $-1 \le X \le 3$.  So in summary we have $$\Pr[Y \le y] = \begin{cases} 0, & y < 0 \\
\Pr[-y^{1/4} \le X \le y^{1/4}], & 0 \le y \le 1 \\
\Pr[-1 \le X \le y^{1/4}], & 1 < y \le 81 \\
1, & y > 81. \end{cases}$$  Now there are no further restrictions and we can proceed with the evaluation of each case.
