Determine the value of $c$ so that the following function is a pdf.
$f(x) = \begin{cases} {15\over64}+{x\over64}&-2\leq x\leq0 \\{3\over8}+cx &{0<x\leq{3}} \\ 0&\text{otherwise}\end{cases}$
The example says that it is obvious $-2$ and $0$ are points at which $f(x)$ is discontinuous. I am not seeing this connection, and it seems pretty important to notice. Can anyone tell me how the function is discontinuous at these points?
Hint: $f(-2) = \frac{13}{64}$ and $f(x)=0$ for all $x<-2$. Thus $f$ is discontinuous at $x=-2$. There is a jump.
Similarly $f$ is discontinuous at $0$. Discontinuity at $3$ depends on $c$.
For $c$, just find it such that $\int_{-2}^{3} f(x) dx = 1.$