Finding PDF/CDF of a function g(x) as a continuous random variable Suppose that $X$ is a continuous random variable with PDF:
$$f_X(x)=\begin{cases}
4x^3 & 0<x\le 1\\
0 & otherwise
\end{cases}$$
and let $Y=\frac{1}{X}$. Find $f_Y(y).$

My approach:
First, find the CDF, $F_X(x)$:
$$F_X(x)=\begin{cases}
0, & x\le0\\
x^4, & 0<x\le 1\\
1 & x>1
\end{cases}
$$
Next, find range of $Y$, $Y\in[1,\infty)$,since $x$ can only take on value of $(0, 1]$, therefor, $\frac{1}{x}$ has range of $[1, \infty)$.
With that in mind, attempt to find CDF $F_Y(y)$, but I am stuck here and not sure how to proceed:
$$F_Y(y)=\begin{cases}
0 &y<1\\
? &1\le y<\infty
\end{cases}
$$
UPDATE 12/16/2022
I got the limit thanks to the tips from @Ben
$$Y=\frac{1}{X}\\
0\le x \le 1 \Rightarrow 0\le \frac{1}{y}\le 1 \\
\frac{1}{y}\le 1 \Rightarrow 1\le y \\
0\le \frac{1}{y} \Rightarrow y\le\frac{1}{0}\Rightarrow y\le\infty \\
\text{therefore, the limit for y is: } 1\le y\le\infty \\
$$
Now to find $f_Y(y)$, I suppose one has to start from CDF
$$F_X(x)=\begin{cases}
0, & x\le0\\
x^4, & 0<x\le 1\\
1 & x>1
\end{cases}\\
\text{ Therefore: }\\
F_Y(y)=\begin{cases}
0, & y\le1\\
\frac{1}{y^4}, & 1\le y\le\infty\\
1 & y\ge\infty\text{ ?? This doesn't make sense}\\
\end{cases}\\
$$

Since $Y=1/X$ you have:
$$\begin{align}
F_Y(y) = \mathbb{P}(Y \leqslant y) 
&= \mathbb{P}(1/X \leqslant y) \\[6pt]
&= \mathbb{P}(X \geqslant 1/y) \\[6pt]
&= 1 - \mathbb{P}(X < 1/y) \\[6pt]
&= 1 - F_X(1/y) \\[6pt]
&= \begin{cases}
0 & & & \text{if } y < 1, \\[6pt]
1-1/y^4 & & & \text{if } y \geqslant 1. \\[6pt]
\end{cases} \\[6pt]
\end{align}$$
(Note that in the transition to the fourth line I have used the fact that $X$ is a continuous random variable so that $\mathbb{P}(X=1/y)=0$, which means that $\mathbb{P}(X < 1/y) = \mathbb{P}(X \leqslant 1/y)$.)  We then get the corresponding density function:
$$f_Y(y) = \frac{4}{y^3} 
\quad \quad \quad \quad \quad 
\text{for } y \geqslant 1.$$
