Question

which of the following accurately lists all discontinuities of the function below?
f(x)=\begin{cases}4, & x < - 4\\(x + 2)^2, &-4leq xleq - 2\\-\frac{1}{2}x + 1, &-2 < x < 4\\-1, &x>4end{cases}
point discontinuity at (x=-2)
point discontinuity at (x = 4); jump discontinuity at (x=-2)
point discontinuities at (x=-4) and (x = 4); jump discontinuity at (x=-2)
jump discontinuities at (x=-4,x=-2,) and (x = 4)

Explanation:

Step1: Check continuity at \(x = - 4\)

Left - hand limit as \(x\to - 4^{+}\):
\(\lim_{x\to - 4^{+}}(x + 2)^{2}=(-4 + 2)^{2}=4\)
Right - hand limit as \(x\to - 4^{-}\): The function is not defined for \(x\lt - 4\), but the value of the function at \(x=-4\) is \(4\). So, the function is continuous at \(x = - 4\).

Step2: Check continuity at \(x=-2\)

Left - hand limit as \(x\to - 2^{-}\): \(\lim_{x\to - 2^{-}}(x + 2)^{2}=0\)
Right - hand limit as \(x\to - 2^{+}\): \(\lim_{x\to - 2^{+}}(-\frac{1}{2}x + 1)=-\frac{1}{2}\times(-2)+1=2\)
Since the left - hand limit and right - hand limit are not equal, there is a jump discontinuity at \(x=-2\).

Step3: Check continuity at \(x = 4\)

Left - hand limit as \(x\to 4^{-}\): \(\lim_{x\to 4^{-}}(-\frac{1}{2}x + 1)=-\frac{1}{2}\times4+1=-1\)
Right - hand limit as \(x\to 4^{+}\): The value of the function for \(x\gt4\) is \(-1\), and the function value at \(x = 4\) is not defined in the piece - wise function for the sub - domain \(-2\lt x\lt4\). So, there is a point discontinuity at \(x = 4\).

Answer:

point discontinuity at \(x = 4\); jump discontinuity at \(x=-2\)