jeanettedavis255
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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]

Sagot :

GinnyAnswer
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].
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