Which of the following accurately lists all discontinuities of the function below?

[tex]\[ f(x) = \left\{ \begin{aligned} 4, & \quad x \ \textless \ -4 \\ (x + 2)^2, & \quad -4 \leq x \leq -2 \\ -\frac{1}{2}x + 1, & \quad -2 \ \textless \ x \ \textless \ 4 \\ -1, & \quad x \ \textgreater \ 4 \end{aligned} \right. \][/tex]

A. Point discontinuity at [tex]$ x = -2 $[/tex]

B. Point discontinuity at [tex]$ x = 4 $[/tex]; jump discontinuity at [tex]$ x = -2 $[/tex]

C. Point discontinuities at [tex]$ x = -4 $[/tex] and [tex]$ x = 4 $[/tex]; jump discontinuity at [tex]$ x = -2 $[/tex]

D. Jump discontinuities at [tex]$ x = -4 $[/tex], [tex]$ x = -2 $[/tex], and [tex]$ x = 4 $[/tex]

Question

10-07-2024
Mathematics

Answered

Which of the following accurately lists all discontinuities of the function below?

[tex]\[ f(x) = \left\{ \begin{aligned} 4, & \quad x \ \textless \ -4 \\ (x + 2)^2, & \quad -4 \leq x \leq -2 \\ -\frac{1}{2}x + 1, & \quad -2 \ \textless \ x \ \textless \ 4 \\ -1, & \quad x \ \textgreater \ 4 \end{aligned} \right. \][/tex]

A. Point discontinuity at [tex]$ x = -2 $[/tex]

B. Point discontinuity at [tex]$ x = 4 $[/tex]; jump discontinuity at [tex]$ x = -2 $[/tex]

C. Point discontinuities at [tex]$ x = -4 $[/tex] and [tex]$ x = 4 $[/tex]; jump discontinuity at [tex]$ x = -2 $[/tex]

D. Jump discontinuities at [tex]$ x = -4 $[/tex], [tex]$ x = -2 $[/tex], and [tex]$ x = 4 $[/tex]

Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2024-07-10T14:02:25-04:00

To determine the discontinuities of the given piecewise function, we need to evaluate the continuity at the transition points, specifically at [tex]$ x = -4 $[/tex], [tex]$ x = -2 $[/tex], and [tex]$ x = 4 $[/tex].

Let's consider each of these points one by one:

1. At [tex]$ x = -4 $[/tex]:
- For [tex]$ x < -4 $[/tex]: [tex]$ f(x) = 4 $[/tex].
- For [tex]$ -4 \leq x \leq -2 $[/tex]: [tex]$ f(x) = (x + 2)^2 $[/tex].
- As [tex]$ x $[/tex] approaches [tex]$-4$[/tex] from the left, [tex]$ f(x) $[/tex] approaches 4.
- As [tex]$ x $[/tex] approaches [tex]$-4$[/tex] from the right, [tex]$ f(x) = (-4 + 2)^2 = 4 $[/tex].

Since both the left-hand limit and right-hand limit are equal and [tex]$ f(-4) = 4 $[/tex], the function is continuous at [tex]$ x = -4 $[/tex].

2. At [tex]$ x = -2 $[/tex]:
- For [tex]$ -4 \leq x \leq -2 $[/tex]: [tex]$ f(x) = (x + 2)^2 $[/tex].
- For [tex]$ -2 < x < 4 $[/tex]: [tex]$ f(x) = -\frac{1}{2} x + 1 $[/tex].
- As [tex]$ x $[/tex] approaches [tex]$-2$[/tex] from the left, [tex]$ f(x) = (-2 + 2)^2 = 0 $[/tex].
- As [tex]$ x $[/tex] approaches [tex]$-2$[/tex] from the right, [tex]$ f(x) = -\frac{1}{2} \cdot (-2) + 1 = 2 $[/tex].

Since the left-hand limit (0) is not equal to the right-hand limit (2) as [tex]$ x $[/tex] approaches [tex]$-2$[/tex], there is a jump discontinuity at [tex]$ x = -2 $[/tex].

3. At [tex]$ x = 4 $[/tex]:
- For [tex]$ -2 < x < 4 $[/tex]: [tex]$ f(x) = -\frac{1}{2} x + 1 $[/tex].
- For [tex]$ x > 4 $[/tex]: [tex]$ f(x) = -1 $[/tex].
- As [tex]$ x $[/tex] approaches \4 from the left, [tex]$ f(x) = -\frac{1}{2} \cdot 4 + 1 = -1 $[/tex].
- As [tex]$ x $[/tex] approaches \4 from the right, [tex]$ f(x) = -1 $[/tex].

Since both the left-hand limit and right-hand limit are equal and [tex]$ f(4) = -1 $[/tex], the function is continuous at [tex]$ x = 4 $[/tex].

Given this analysis, the function [tex]$ f(x) $[/tex] has a jump discontinuity at [tex]$ x = -2 $[/tex].

Therefore, the accurate listing of all discontinuities of the function is:

- Jump discontinuity at [tex]$ x = -2 $[/tex]

Hence, the correct option is:
- jump discontinuity at [tex]$ x = -2 $[/tex].