Question

which of the following accurately lists all discontinuities of the function below?
$f(x)=\

$$\begin{cases}4, & x < -4 \\\\ (x + 2)^2, & -4 \\leq x \\leq -2 \\\\ -\\frac{1}{2}x + 1, & -2 < x < 4 \\\\ -1, & x > 4 \\end{cases}$$

$
\bigcirc point discontinuity at $x = -2$
\bigcirc point discontinuity at $x = 4$; jump discontinuity at $x = -2$
\bigcirc point discontinuities at $x = -4$ and $x = 4$; jump discontinuity at $x = -2$
\bigcirc jump discontinuities at $x = -4$, $x = -2$, and $x = 4$

Explanation:

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

• Left - hand limit (LHL) as \( x\to - 4^{-}\): For \( x < - 4\), \( f(x)=4 \), so \( \lim_{x\to - 4^{-}}f(x)=4 \)
• Right - hand limit (RHL) as \( x\to - 4^{+}\): For \( - 4\leq x\leq - 2\), \( f(x)=(x + 2)^{2} \), substitute \( x=-4 \), we get \( (-4 + 2)^{2}=(-2)^{2}=4 \)
• \( f(-4)=( - 4+2)^{2}=4 \)
• Since \( \lim_{x\to - 4^{-}}f(x)=\lim_{x\to - 4^{+}}f(x)=f(-4) = 4 \), the function is continuous at \( x=-4 \)

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

• LHL as \( x\to - 2^{-}\): For \( - 4\leq x\leq - 2\), \( f(x)=(x + 2)^{2} \), substitute \( x=-2 \), we get \( (-2 + 2)^{2}=0 \)
• RHL as \( x\to - 2^{+}\): For \( - 2
• Since \( \lim_{x\to - 2^{-}}f(x)=0

eq\lim_{x\to - 2^{+}}f(x)=2 \), there is a jump discontinuity at \( x=-2 \)

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

• LHL as \( x\to 4^{-}\): For \( - 2
• RHL as \( x\to 4^{+}\): For \( x>4\), \( f(x)=-1 \), so \( \lim_{x\to 4^{+}}f(x)=-1 \)
• The function is not defined at \( x = 4 \) (the left - hand side is for \( x<4 \) and the right - hand side is for \( x>4 \), there is no definition of \( f(4) \) in the piecewise function). But \( \lim_{x\to 4^{-}}f(x)=\lim_{x\to 4^{+}}f(x)=-1 \), but since \( f(4) \) is not defined, it is a point discontinuity (removable discontinuity, but in terms of the options, it is a point discontinuity)

Now let's analyze the options:

• Option 1: Only says point discontinuity at \( x=-2 \), but we have a point discontinuity at \( x = 4 \) and jump at \( x=-2 \), so incorrect.
• Option 2: Point discontinuity at \( x = 4 \) (since \( f(4) \) is not defined but limits exist) and jump discontinuity at \( x=-2 \) (since left and right limits exist but are not equal), this matches our analysis.
• Option 3: Says point discontinuities at \( x=-4 \) and \( x = 4 \), but \( x=-4 \) is continuous, so incorrect.
• Option 4: Says jump discontinuities at \( x=-4,x=-2,x = 4 \), but \( x=-4 \) is continuous and \( x = 4 \) is a point discontinuity, so incorrect.

Answer:

point discontinuity at \( x = 4 \); jump discontinuity at \( x = -2 \) (the option corresponding to this is the second option: point discontinuity at \( x = 4 \); jump discontinuity at \( x = -2 \))