$f_x = \begin{cases} 0 & \text{else } \\ \frac{1}{9}\left(3-|x|\right) & \text{ } |x| \le 3 \end{cases}$
Now I know the CDF is defined as:
$\int_\infty^af_xdx$
and I believe I need to perform the following integrals:
for $x\lt3 \rightarrow0$
$\int_{-3}^{x}f_xdx$ for $-3 \le x \lt 0$
$\int_{x}^{3}f_xdx$ for $0 \le x \le 3$
and
$\int^{\infty}_3f_x dx $
But my answers don't seem to be correct, leading me to believe that my limits of integration are incorrect, or that the absolute value plays a much bigger when integrating on each interval.
For the CDF $F(x)=\int_{-\infty}^x f_t(t) dt\;$ you have $$F(x) = 0, \quad x\le -3$$ $$F(x)= \int_{-3}^x \frac{1}{9}(3+x) dt= \frac{1}{2}+ \frac{1}{3}x+ \frac{1}{18}x^2, \quad -3<x<0$$ $$F(x)= \int_{-3}^x \frac{1}{9}(3-|x|) dt= \int_{-3}^0 \frac{1}{9}(3+x) dt + \int_{0}^x \frac{1}{9}(3-x) = \frac{1}{2} + \frac{1}{3}x- \frac{1}{18}x^2,\quad 0\le x <3$$ $$F(x) = 1, \quad x\ge 3$$