jeanettedavis255
Answered

Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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].
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.