which of the following accurately lists all discontinuities of the function below?\nf(x)=\begin{cases}4, & x…

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

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

Answer

Answer:

D. jump discontinuities at (x = - 4,x=-2,) and (x = 4)

Explanation:

Step1: Check (x=-4)

Left - hand limit: (\lim_{x\rightarrow - 4^{-}}f(x)=4). Right - hand limit: (\lim_{x\rightarrow - 4^{+}}f(x)=( - 4 + 2)^{2}=4). But function value at (x=-4) is not defined in the first piece and ((-4 + 2)^{2}=4) in the second piece. There is a jump as the function changes rules.

Step2: Check (x=-2)

Left - hand limit: (\lim_{x\rightarrow - 2^{-}}f(x)=(-2 + 2)^{2}=0). Right - hand limit: (\lim_{x\rightarrow - 2^{+}}f(x)=-\frac{1}{2}\times(-2)+1=2). There is a jump discontinuity.

Step3: Check (x = 4)

Left - hand limit: (\lim_{x\rightarrow4^{-}}f(x)=-\frac{1}{2}\times4 + 1=-1). Right - hand limit: (\lim_{x\rightarrow4^{+}}f(x)=-1). But the function changes rules here, and there is a jump as the way the function is defined changes.