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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

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?

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    Actually the dis/continuity of the PDF at these points is peripheral and quite unnecessary to solve the question. Simply put: just forget it. – 2017-02-28
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    @Did The next part asked for $P(-1\leq X\leq 1)$ and I was able to disregard it for that as well. In the 3rd part it wants the CDF, or F(x). Can I disregard it for this as well? – 2017-02-28
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    It is discontinuous at $3$ too. But as @Did mentioned, these aren't required to evaluate $c$! – 2017-02-28
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    Yes, this is irrelevant for the CDF as well. – 2017-02-28
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    @stud_iisc can I ask how you can notice this though? I feel that even though it is irrelevant here, it may come in handy later to know how to recognize when it is discontinuous. – 2017-02-28

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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 $\int_{-2}^{3} f(x) dx = 1.$