Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. The Statement of the Problem:
Let $X$ have pdf
$$f_X(x) =
\begin{cases}
\frac{1}{4}  & 0<x<1 \\
\frac{3}{8}  & 3<x<5 \\
0            & \text{otherwise}
\end{cases}$$
(a) Find the cumulative distribution function of $X.$
(b) Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. Hint: Consider three cases: $1/5 \le y \le 1/3, 1/3 \le y \le 1,$ and $ y \ge 1.$
Where I Am:
I think I did part (a) correctly. I did the following:
$$F_X(x) =
\begin{cases}
\frac{1}{4}x +c_1  & 0<x<1 \\
\frac{3}{8}x + c_2  & 3<x<5 \\
0            & \text{otherwise}
\end{cases}$$
$$F(0)=0=\frac{1}{4}(0)+c_1 \implies c_1 = 0 $$
$$F(5)=1=\frac{3}{8}(5)+c_2 \implies c_2 = -\frac{7}{8} $$
Therefore:
$$F_X(x) =
\begin{cases}
\frac{1}{4}x   & 0<x<1 \\
\frac{1}{4}    & 1<x<3 \\
\frac{3}{8}x - \frac{7}{8}  & 3<x<5 \\
1            & x > 5
\end{cases}$$
If that's not right, however, please let me know.
Now, for part (b), I got a little lost. Here's what I did:
$$ \text{Let } g(x) = \frac{1}{x} \implies g'(x) = -\frac{1}{x^2} $$
Then:
$$ f_Y(y)=\frac{f_X(x)}{\lvert g'(x) \rvert} = \frac{f_X(\frac{1}{y})}{\lvert g'(\frac{1}{y})\rvert} $$
Therefore:
$$f_Y(y) =
\begin{cases}
\frac{1}{4}  & 0<\frac{1}{y}<1 \\
\frac{3}{8}  & 3<\frac{1}{y}<5 \\
0            & \text{otherwise}
\end{cases}$$
and taking reciprocals and flipping inequalities...
$$f_Y(y) =
\begin{cases}
\frac{1}{4}  & y \ge 1 \\
\frac{3}{8}  & \frac{1}{5} \le y \le \frac{1}{3} \\
0            & \frac{1}{3} \le y \le 1
\end{cases}$$
This, however... doesn't seem right. For example, what is $f_Y(y)$ when $y \in [0, \frac{1}{5}]$? Is it just $0$? I know I did something wrong here, but I can't quite figure out what exactly. If anybody could help me out, I'd appreciate it.
EDIT: "Second" Attempt...
$$F_X \left( \frac{1}{y} \right) =
\begin{cases}
\frac{1}{4}\left( \frac{1}{y} \right)   & y \ge 1 \\
\frac{1}{4}    & \frac{1}{3} \le y \le 1 \\
\frac{3}{8}\left( \frac{1}{y} \right) - \frac{7}{8}  & \frac{1}{5} \le y  \le \frac{1}{3} \\
0            & \text{otherwise}
\end{cases}$$
Therefore:
$$ f_Y(y)=\frac{d}{dy}F_X \left( \frac{1}{y} \right)= \begin{cases}
-\frac{1}{4}\left( \frac{1}{y^2} \right)   & y \ge 1 \\
-\frac{3}{8}\left( \frac{1}{y^2} \right)  & \frac{1}{5} \le y  \le \frac{1}{3} \\
0            & \text{otherwise}
\end{cases} $$

$$
f_X(x) =
\begin{cases}
\frac{1}{4}  & 0<x<1 \\
\frac{3}{8}  & 3<x<5 \\
0            & \text{otherwise}
\end{cases}
$$
\begin{align}
f_Y(y) & = \frac d {dy} F_Y(y) = \frac d {dy} \Pr(Y\le y) = \frac d {dy} \Pr\left( \frac 1 X \le y \right) \\[10pt]
& = \frac d {dy} \Pr\left( X\ge \frac 1 y \right) \text{ (if }y>0) \\[10pt]
& = \frac d {dy} F_X\left(\frac 1 y \right).
\end{align}
Now notice that when $3<x<5$ then $\dfrac 1 5 <\dfrac 1 x < \dfrac 1 3$, or $\dfrac 1 5 < y < \dfrac 1 3$ and similarly for other intervals.
In part $(a)$ I'd use definite integrals.  Since $f_X(x)=0$ when $x<0$, you have for $x\ge 0$,
$$
F_X(x) = \int_0^x f_X(w) \, dw.
$$
If $x>1$, this becomes
$$
F_X(x) = \int_0^x f_X(w) \, dw = \int_0^1 f_X(w)\,dw + \int_1^x f_X(w)\,dw
$$
and if $x>3$ then
$$
F_X(x) = \int_0^x f_X(w) \, dw = \int_0^1 f_X(w)\,dw + \int_1^3 f_X(w)\,dw + \int_3^x f_X(w)\,dw.
$$
