I'm sure I solved this correctly, but the bounds are a bit tricky.
There's an rv w/pdf $$ f(y) = \frac{a+by}{c} $$ bounded on $(1,3)$ and function $W=Y^2$, I need to compute $f(w)$.
I decided to take a longer route, and start with CDF of $Y$, which comes out to
$$
F(y) = \frac{a}{c}(y-1) + \frac{b}{2c}(y^2 - 1)
$$
and, since $Y$ is bounded, and $W=Y^2, W$ is also bounded: $(1,9)$. Now, using the CDF approach,
$$
P(W<w)= P(Y^2<w) = P(-\sqrt{w}<Y<\sqrt{w}) = F_{Y}(\sqrt{w})-F_{Y}(-\sqrt{w})
$$
This is where my confusion started, becasue $-\sqrt{w}$ is certainly negative, and $F(y)$ is only defined for $y>1$. What should the lower bound be then? I used $F(1)=0$ to get
$$
F_W(w) = \frac{a}{c}(\sqrt{w}-1) + \frac{b}{2c}(w-1), 1 <w<9
$$
Differentiating wrt $w$, I get
$$
f_W(w) = \frac{a}{2c \sqrt{w}} + \frac{b}{2c}
$$
which should be the correct answer, but I'm still hesitant about that lower bound $F_{Y}(1)$. Is it correct?
Sorry but
$$f_Y(y)=\int_1^3\frac{a+by}{c}dy \ne 1$$
thus you have to normalize $f_Y$ first...
The problem is that I try to normalize it I get
$$f_Y(y)=\frac{a+by}{2(2b+a)}\cdot\mathbb{1}_{(1;3)}(y)$$
thus $c$ disappears...
So check you $f_Y$ first...I can assume that $a,b$ are density parameters while $c$ is a constant to be derived. In this case, using the following formula
$$f_W(w)=f_Y[g^{-1}(w)]\cdot|\frac{d}{dw}g^{-1}(w)|$$
You immediately get your desired density
$$f_W(w)=\frac{a+b\sqrt{w}}{4(2b+a)\sqrt{w}}\cdot\mathbb{1}_{(1;9)}(w)$$