to find \(f_{Y}(y)\) given \(f_{X}(x)\) and \(Y=g\left ( X\right ) \) , divide the \(g\left ( X\right ) \) region into 3 parts:
part 1: \(1.5\leq x\leq 2\)
part 2: \(.5\leq x<1.5\)
part 3: \(0\leq x<.5\)
and then use the fundemental theorm of probabilities (page 93 of notes), which says:
\[ f_{Y}\left ( y\right ) =\frac{f_{X}(x_{1})}{\left | g^{^{\prime }}\left ( X_{1}\right ) \right | }+\frac{f_{X}(x_{2})}{\left | g^{^{\prime }}\left ( X_{2}\right ) \right | }+\cdot \cdot \cdot +\frac{f_{X}(x_{n})}{\left | g^{^{\prime }}\left ( X_{n}\right ) \right | } \]
where \(n\) is the number of parts over which region \(g\left ( X\right ) \) was divided, here \(n=3\;\)and \(f_{X}\left ( x\right ) =\frac{3x}2-\frac{3x^{2}}4\) over \(0\leq x\leq 2\) and \(0\,\)everywhere else.
part1:
\(x_{1}\equiv 1.5\leq x\leq 2\Longrightarrow -1\leq y\leq 0\)
\[ g\left ( X_{1}\right ) =2x-4=y \]
so \begin{equation} \label{1}x=\frac{y+4}2 \end{equation}
now\[ g^{^{\prime }}\left ( X_{1}\right ) =2 \]
so \begin{equation} \label{2}f_{Y}\left ( y\right ) =\frac{f\left ( x_{1}\right ) }{\left | g\left ( X_{1}\right ) \right | }=\frac{\frac{3x}2-\frac{3x^{2}}4}2=3x-\frac{3x^{2}}2 \end{equation}
from 0.1 and 0.2 we get\[\begin{array} [t]{l}f_{Y}\left ( y\right ) =3x- \frac{3x^{2}}2\\ \\ \;\;=3\left ( \frac{y+4}2\right ) -\frac 32\left ( \frac{y+4}2\right ) ^{2}\\ \\ \;\;=\frac 32\left ( y+4\right ) -\frac 38\left ( y+4\right ) ^{2}\\ \\ \;\;=\frac 32y+6-\frac 38\left ( y^{2}+16+8y\right ) \\ \\ \;\;=\frac 32y+6-\frac 38y^{2}-6-3y\\ \\ \;\;=-\frac 38y^{2}-\frac 32y \end{array} \]
so over \(-1\leq y\leq 0\)\[ f_{Y}\left ( y\right ) =-\frac 38y^{2}-\frac 32y \]
part2:
\(x_{2}\equiv 0.5\leq x<1.5\Longrightarrow y=0\)
over this part, since \(g\left ( X_{2}\right ) =0\) then \(f_{Y}\left ( y\right ) \) is an impulse
\[ f_{Y}\left ( y\right ) =\frac{f\left ( x_{2}\right ) }{\left | g^{^{\prime }}\left ( X_{2}\right ) \right | }=P\left ( .5\leq x\leq 1.5\right ) \delta \left ( y\right ) \]
but \[\begin{array} [c]{c}P\left ( .5\leq x\leq 1.5\right ) =F_{X}\left ( 1.5\right ) -F_{X}\left ( .5\right ) \\ \\ =\int _{-\infty }^{1.5}f_{X}\left ( x\right ) \;dx\;-\int _{-\infty }^{.5}f_{X}\left ( x\right ) \;dx\\ \\ =\int _{0}^{1.5} \frac{3x}2-\frac{3x^{2}}4\;dx-\int _{0}^{.5}\frac{3x}2-\frac{3x^{2}}4dx\\ \\ =0.84375-0.15625\\ \\ =0.6875 \end{array} \]
so at \(y=0\)\[ f_{Y}\left ( y\right ) =0.6875\;\delta \left ( y\right ) \]
part3:
\(x_{3}\equiv 0\leq x<0.5\Longrightarrow 0\leq y\leq 1\)
\[ g\left ( X_{3}\right ) =-2x+1=y \]
so \begin{equation} \label{3}x=\frac{1-y}2 \end{equation}
now\[ g^{^{\prime }}\left ( X_{3}\right ) =-2 \]
so \begin{equation} \label{4}f_{Y}\left ( y\right ) =\frac{f\left ( x_{3}\right ) }{\left | g\left ( X_{3}\right ) \right | }=\frac{\frac{3x}2-\frac{3x^{2}}4}2=3x-\frac{3x^{2}}2 \end{equation}
from 0.3 and 0.4 we get\[\begin{array} [t]{l}f_{Y}\left ( y\right ) =3x- \frac{3x^{2}}2\\ \\ \;\;=3\left ( \frac{1-y}2\right ) -\frac 32\left ( \frac{1-y}2\right ) ^{2}\\ \\ \;\;=\frac 32\left ( 1-y\right ) -\frac 38\left ( 1-y\right ) ^{2}\\ \\ \;\;=\frac 32-\frac 32y-\frac 38\left ( y^{2}+1-2y\right ) \\ \\ \;\;=\frac 32-\frac 32y-\frac 38y^{2}-\frac 38+\frac 34y\\ \\ \;\;=-\frac 38y^{2}-\frac 34y+\frac 98 \end{array} \]
so over \(0\leq y\leq 1\)\[ f_{Y}\left ( y\right ) =-\frac 38y^{2}-\frac 34y+\frac 98 \]