Answer:
B. (-1, 2)
Step-by-step explanation:
A piecewise-defined function may be discontinuous at the boundaries of its pieces, and it may be discontinuous if it is undefined anywhere within one of its pieces. So, you need to check all of these possibilities.
The boundary locations are x = -1, 0, and 4.
At x=-1, the function has the value e^-1 on one side of the boundary and -1 on the other side, so is discontinuous at x=-1. This rules out choice A.
At x=0, the function has the value 0 on both sides of the boundary, so it is continuous there.
At x=4, the function has the value 10 on one side of the boundary and the value cos(12) on the other side, so is discontinuous at x=4.
In the piece between 0 and 4, the function is defined as 5x/(x-2), so will be undefined where the denominator is zero, at x=2.
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In summary, the function is discontinuous at x=-1, x=2, x=4. So, any interval containing these x-values can be ruled out. The only remaining possibility is ...
B. (-1, 2)
