If $X$ is uniformly distributed $\mathcal{U}(0,1)$ , then what is the distribute density function of $Y$? I thought that if $$fx(x) = 1/(1-0), \; \mbox{for} \; 0<x<1$$ then $$fy(Y=2x-4)=fx((y+4)/2) = 1(1-0), \; \mbox{for} \; -2<(y+4)/2<2$$
Am I right?
On
Hint:
If $X$ is uniformly distributed and $a,b\in\mathbb R$ with $a\neq0$ then $Y:=aX+b$ is also uniformly distributed. This in the meaning that its PDF has a positive constant value on a certain set and takes value $0$ on the complement of that set.
If you know that the PDF of $Y$ takes constant positive value $c$ on e.g. some interval $(p,q)$ and takes value $0$ otherwise then you can find the determining constant on base of $$1=\int f_Y(y)dy=\int_{(p,q)}cdy=c(q-p)$$ In your case $X$ takes values in $(0,1)$ so that $2X-4$ takes values in $(-4,-2)$ leading to $1=c((-2)-(-4))=2c$ hence $c=\frac12$.
On
For a random variable $ X $ distributed according to the pdf $ p_X $ and a differentiable function $ f $ with positive derivative, you can show that the random variable $ f(X) $ is distributed according to $$ p_{f(X)}(y) = \frac{p_X(f^{-1}(y))}{f^\prime(f^{-1}(y))}. $$
In your case, you have that $ p_X(x) = \chi_{[0, 1]}(x) $ (with $ \chi_{A} $ the indicator function of set $ A $), $ f^{-1}(y) = (y + 4)/2 $ and $ f^\prime(y) = 2 $, hence $$ p_{f(X)}(y) = \chi_{[-4, -2]}(y)/2 $$ (since $ \chi_{[0, 1]}((y + 4)/2) = \chi_{[-4, -2]}(y) $).
No:
The cdf of $2X-4$ is $$F_{2X-4}(x)=P(2X-4<x)=P\left(X<\frac{x+4}2\right)= \begin{cases} 0,& \text{ if } x<-4\\ \frac{x+4}2,& \text{ if } -4\le x\le -2\\ 1,&\text{ if } x>-2 \end{cases}.$$
The density is $$f_{2X-4}(x)=\frac{dF_{2X-4}(x)}{dx}= \begin{cases} \frac 12,& \text{ if } -4\le x\le -2\\ 0,&\text{ otherwise } \end{cases}.$$