Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of $\displaystyle Y=X^{2}$?
You need to be careful.
This one to one on 1<x<3 and two to one -1<x<1.
USE $\displaystyle f_Y(y)= \biggl(f_X(-\sqrt y) +f_X(\sqrt y)\biggr)\biggl({1\over 2\sqrt y}\biggr)$.
Now $\displaystyle f_X(x)={1\over 4}$ on -1<x<3.
Next notice that y's range is 0 to 9.
WHEN 0<y<1, we have a 2-1 mapping and both
$\displaystyle f_X(\sqrt y)=f_X(-\sqrt y)=1/4$.
So, for 0<y<1, we have
$\displaystyle f_Y(y)= \biggl( {1\over 4} +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 4\sqrt y}$.
But when 1<y<9, we have a 1-1 mapping (with probability one)
So, $\displaystyle f_X(\sqrt y)=1/4$ while $\displaystyle f_X(-\sqrt y)=0$.
Thus for 1<y<9, we have
$\displaystyle f_Y(y)= \biggl( 0 +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 8\sqrt y}$.
AND zero everwhere else.