An exercise applying "The method of Distribution Functions" I'm not understanding something about this question:


A process for refining sugar yields up to 1 ton of sugar per day, but the actual amount produced, $Y$ is a random variable because of the machine breaking down and other slow downs. Suppose that $Y$ has density function given by:
    $$f(y)=\begin{cases}
2y&\text{,}& 0\leq y \leq 1\\
0&\text{ otherwise. }
\end{cases}$$
    The company is paid at a rate of \$300 per ton for the sugar, but it also has fixed overhead costs of \$100 per day. The the daily profit, in hundreds of dollars, is $U = 3Y -1$. Find the probability density function of $U$.  


So anyway, they go on to say that:
$$F_{u}(u)= P(3Y-1\leq u) = P(Y \leq \frac{u+1}{3})$$
and from this:


if $u < -1$, then $(u+1)/3<0$ and, therefore, $F_u(u)=P(Y\leq\frac{u+1}{3})=0$  


Now this I understand, however, I do not know why they come to this next part, and I would like to know: 


also if $u>2$, then $\frac{(u+1)}{3}>1$ and $F_u(u)=P(Y \leq \frac{u+1}{3})=1$.  


I thought that would only be the case if the original pdf is but I would not be surprised if I'm not understanding something:
$$f(y)=\begin{cases}
y&\text{,}& 0\leq y \leq 1\\
0&\text{ otherwise. }
\end{cases}$$  
Can anyone help me with this one?   

Because $f_Y(y) = \begin{cases}2y &:& {0\leqslant y\leqslant 1}\\0&:& \text{elsewhere}\end{cases}$, therefore: $$F_Y(y) = \begin{cases} 0 &:& \qquad y < 0 \\ y^2 &:& ~0\leqslant y < 1\\ 1 &:& ~1\leqslant y\end{cases}$$
Then we have that for $g(u)=\dfrac{u+1}{3}$.
$$\begin{align} F_U(u) &= F_Y(g(u))\\[1ex] & =\begin{cases} 0 &:& \qquad g(u)<0\\ g^2(u) &:& ~0\leqslant g(u)< 1\\ 1 &:& ~1\leqslant g(u)\end{cases}\\[1ex] &= \begin{cases}0 &:& \qquad\qquad~~~ u< 3(0)-1\\ (u+1)^2/9&:& 3(0)-1\leqslant u< 3(1)-1\\ 1&:& 3(1)-1\leqslant u\end{cases}\\[1ex] &= \begin{cases}0 &:& \qquad~~ u< -1\\ (u+1)^2/9&:& -1\leqslant u< ~~~2\\ 1&:& ~~~2\leqslant u\end{cases} \end{align}$$
Then, of course $$f_U(u)=\begin{cases}2(u+1)/9 &:& -1\leqslant u< 2\\ 0 &:& \text{elsewhere}\end{cases}$$
