Find the PDF of $Y = X^4 + 1$ if $X\sim\exp(\lambda)$.
When a transformation is not one-to-one, we have multiple solutions for $X$. Take for example $Y = X^2$. Then \begin{align*} x_1 &= \sqrt{y}\\ x_2 &= -\sqrt{y} \end{align*} Now $Y = X^4 + 1$ isn't one-to-one since $Y = 2$ when $X= \pm 1, \pm i$. However, I don't see what my $x_i$s would be for $X = \sqrt[4]{Y - 1}$.
$$\begin{align*}F_Y(y)&=P\left(X^4+1\le y\right)=P\left(|X|\le \sqrt[4]{y-1}\right)=P\left(-\sqrt[4]{y-1}\le X\le \sqrt[4]{y-1}\right)=\\&=F_X\left(\sqrt[4]{y-1}\right)-F_X\left(-\sqrt[4]{y-1}\right)\end{align*}$$ but since $F_X(x)=0$, for $x<0$ the above reduces to $$F_Y(y)=F_X\left(\sqrt[4]{y-1}\right)$$ for $y\ge1$. Hence $$f_Y(y)=f_X\left(\sqrt[4]{y-1}\right)\cdot\left(\sqrt[4]{y-1}\right)'=λe^{-λ\sqrt[4]{y-1}}\cdot\frac{1}{4\sqrt[4]{(y-1)^3}}$$ for $y> 1$.