Finding $\text{E}(f(x))$,$\text{Var}(f(x))$ and the PDF of $f(x)$ The problem reads, suppose $X\sim N(5,2)$ and $Y=2X+4$.
find $\text{E}(Y)$, $\text{Var}(Y)$ and the PDF $p_Y(y)$.
My attempt. 
$\text{E}[Y]= \text{E}[2X+4]=\text{E}[2X]+\text{E}[4]= 2\text{E}[x]+4$, now since $\mu_X=5$, then $\text{E}[X]=2\cdot 5+4=14$?
$\text{Var}(Y)= \text{Var}[2X+4]=4\text{Var}[x]$, now since $\sigma_X^2=2$, then we get $4\cdot 2=8$?
For $p_Y(y)$ I'm not exactly sure where to start. 
 A: You've done a great job so far, now let's look at

$$p_Y(y) ={d\over dy}P(Y\le y)={d\over dy}P(2X+4\le y) = {d\over dy}P(X\le (y-4)/2)$$

But this is just

$${d\over dy}\int_{-\infty}^{(y-4)/2}p_X(t)\,dt=p_X\left({y-4\over 2}\right).$$

Now plug into what you already know about $p_X(x)$ and you're done!
A: 
Theorem: Let $X$ be a random variable with probability density $f$. Let $Y=r(X)$ where $r$ is a strictly monotone (smooth) function. Then, the probability density $g$ of $Y$ is given by $$g(y)=f(r^{-1}(y))\left|\frac{d}{dy}r^{-1}(y)\right|.$$ See http://www.randomservices.org/random/dist/Transformations.html for details.

In your case, $r(x)=2x+4$ and $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}$.
Can you do the rest?
A: For expectation, linearity rules work
$E(Y) = 2E(X) + 4 = 2(5) + 4 = 14$
$Var(Y) = 4Var(x) = 4*2 = 8$
For the pdf, you want to different the cdf, easily done through FTC 2
CDF = $P(Y \le y)$, from the $X$ variable you see that $Y = 2X + 4$, thus
$P(Y \le y) = P(2X \le y - 4) = P(X \le \frac{y - 4}{2})$.
Can you finish from here? (Only plugging in is left!)
