Before starting the solution recall that in order for a function to be continuous at \(x = a\) both \(f\left( a \right)\) and \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) must exist and we must have,

\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]

Using this idea it should be fairly clear where the function is not continuous.

First notice that at \(x = - 4\) we have,

\[\mathop {\lim }\limits_{x \to - {4^ - }} f\left( x \right) = 3 \ne - 2 = \mathop {\lim }\limits_{x \to - {4^ + }} f\left( x \right)\]

and therefore, we also know that \(\mathop {\lim }\limits_{x \to - 4} f\left( x \right)\) doesn’t exist. We can therefore conclude that \(f\left( x \right)\) is discontinuous at \(x = - 4\) because the limit does not exist.

Likewise, at \(x = 2\) we have,

\[\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = - 1 \ne 5 = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\]

and therefore, we also know that \[\mathop {\lim }\limits_{x \to 2} f\left( x \right)\] doesn’t exist. So again, because the limit does not exist, we can see that \(f\left( x \right)\) is discontinuous at \(x = 2\).

Finally let’s take a look at \(x = 4\). Here we can see that,

\[\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right) = 2 = \mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)\]

and therefore, we also know that \(\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 2\). However, we can also see that \(f\left( 4 \right)\) doesn’t exist and so once again \(f\left( x \right)\) is discontinuous at \(x = 4\) because this time the function does not exist at \(x = 4\).

All other points on this graph will have both the function and limit exist and we’ll have \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\) and so will be continuous.

In summary then the points of discontinuity for this graph are : \(x = - 4\), \(x = 2\) and \(x = 4\).