Finding CDF of $Z$ with PDF of $X$

Question

Suppose the random variable $X$ has PDF
$$f_X(x) = 2x^2~, \quad\text{when}~ 0 < x < 1~,$$ and zero otherwise.
Suppose $Z = 4X+9$, what is the CDF of $Z$? Then find the PDF of $Z$.

I began by finding $X=(Z-9)/4$, and then finding the integral of $f_X(x)$. I'm not sure of next steps.

Answer 1

I began by finding X=(Z-9)/4, and then finding the integral of fx(x). I'm not sure of next steps.

That will give you the CDF of $Z$.

To find the pdf, differentiate with respect to $z$.$$\begin{align}{F}_Z(z)&={F}_X\left[(z-9)/4\right]\\[1ex]&=\int_{-\infty}^{(z-9)/4} f_X(x)~\mathrm d x\\[1ex]&=\mathbf 1_{0\leq (z-9)\le 4}~\int_0^{(z-9)/4} \require{cancel}\cancelto3{\color{red}2}x^2~\mathrm d x+\mathbf 1_{4\leq (z-9)}\\[1ex]&=\lower{2ex}\ddots\\[3ex] {f}_Z(x)&=\dfrac{\mathrm d~~}{\mathrm d z}F_Z(z)\\[1ex]&=\lower{2ex}\ddots\end{align}$$

Answer 2

$P(Z\le z)=P(4X+9\le z)=P(X\le \frac{z-9}{4})=(\frac{z-9}{4})^3$ (the CDF of Z) for $9\le z \le 13$ The CDF $=0$ for smaller $z$ and $=1$ for larger $z$.

The PDF of $z$ is $\frac{3}{4}(\frac{z-9}{4})^2$ for $9\le z \le 13$ and $=0$ otherwise.