Find pdf given transformation I'm given that the random variable $X$ is distributed as $UNIF(0,1)$.
Moreover, I'm given that:
$U = X(1-X)$
Using the CDF method (i.e. finding the CDF, then taking the derivative), how would I find the PDF of $U$? I can't isolate X in the equation, so how would I go about getting the PDF of $U$ as a result? Thank you

Let $F(u) = \Pr(U \le u)$. The minimum and maximum values of $U$ are $0$ and $1/4$, respectively. Therefore, $F(u)=0$ when $u \le 0$, and  $F(u)=1$ when $u \ge 1/4$. 
Below is the calculations for $0 \le u \le 1/4$. Observe that
$$F(u) = \Pr(X(1-X)\le u) = \Pr(X-X^2 \le u) = \Pr(X-X^2-1/4 \le u-1/4).$$
But $-(X-1/2)^2 = -(X^2-X+1/4)=X-X^2-1/4$. Therefore,
$$F(u) = \Pr\left(-\left(X-\frac{1}{2}\right)^2 \le u-\frac{1}{4}\right) 
\\
= \Pr\left(\left(X-\frac{1}{2}\right)^2 \ge \frac{1}{4}-u\right) \\
= \Pr\left(\left|X-\frac{1}{2}\right| \ge \sqrt{\frac{1}{4}-u}\right).$$
Consequently,
$$F(u) = \Pr\left(X \le \frac{1}{2} - \sqrt{\frac{1}{4}-u}\right) + \Pr\left(X \ge \frac{1}{2} + \sqrt{\frac{1}{4}-u}\right).$$
But $\Pr(X \le x)=x$ and $\Pr(X \ge x) = 1-x$. So
$$F(u) = 1-\sqrt{1-4u}.$$
