Find the CDF and PDF of $W$ where $W =.7 - b(.7 - y)^2, 0
A random variable $Y \sim U[0,1]$. Let $W = \frac7{10} - b\cdot(\frac7{10} - y)^2, 0<b<1$.Completely specify the CDF and PDF of $W$.Also show that the PDF of W integrates to $1$.
So I have worked through most of this problem. Importantly, the transformation is $2:1$ on the interval y $\in (\frac25,1]$ and $1:1$ on the interval y $\in [0, \frac25)$. Thus I think the CDF is:
$$\begin{cases}
\frac3{10}+\sqrt{\frac{7/10 - w}b } & \text{for}~~  w \in (\frac7{10} - \frac{49b}{100} ,\, \frac7{10} - \frac{9b}{100}) \\
 2\sqrt{ \frac{7/10 - w}b } & \text{for}~~ w \in (\frac7{10} - \frac{9b}{100} ,\, \frac7{10}) \\
\end{cases}$$
Can anyone confirm this? If I know this is right, I can figure out the rest on my own. When integrating the PDF, I am running into some strange (${}+{}$ or ${}-{}$) situations that integrate to $1$ only if I select a certain sign, so I am having trouble verifying the CDF is correct by integrating the PDF over its support. 

That is not quite what I get...
$$\begin{align}F_W(w) &=\mathsf P(0.7 -b(0.7-Y)^2\leqslant w)
\\[1ex]&=\mathsf P(Y\leq 0.7-\sqrt{\tfrac{0.7-w}b})+\mathsf P (Y\geq 0.7+\sqrt{\tfrac{0.7-w}b})
\\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7-\sqrt{\tfrac{0.7-w}b})\leq 1}+(0.3-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7+\sqrt{\tfrac{0.7-w}b})\leq 1}+\mathbf 1_{0.7<w}
\\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.49b\leq w\leq 0.7}+(0.3-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.09b\leq w\leq 0.7}+\mathbf 1_{0.7\lt w}
\\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.49b\leq w\lt 0.7-0.09b}+(1-2\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.09b\leq w\lt 0.7}+\mathbf 1_{0.7\leq w}
\\[3ex] F_W(w) &= \begin{cases}0&:& \quad w\lt 0.7-0.49b\\0.7-\sqrt{(0.7-w)/b~}&:&0.7-\sqrt{(0.7-w)/b~}\leqslant w\lt 0.7-0.09b\\1-2\sqrt{(0.7-w)/b~}&:& 0.7-0.09b\leqslant w< 0.7\\1&:& 0.7\leqslant w  \end{cases}
\end{align}$$ 
To check, I have $F_W([0.7-0.49b]^-)=0$ and $F_W([0.7]^+)=1$ as we require.
