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Determine for what x-value(s) f(x) shown below is discontinuous. 22 + 2x x < 1 x² – 4 f(x) = -1 x = 1 3 – 4e"-1 x >1 O x = -2 and x 2 only O x = 2 only O x = 1 only O x = -2, 1, and 2 only O x = -2 only

Determine for what x-value(s) f(x) shown below is discontinuous. 22 + 2x x < 1 x² – 4 f(x) = -1 x = 1 3 – 4e"-1 x >1 O x = -2 and x 2 only O x = 2 only O x = 1 only O x = -2, 1, and 2 only O x = -2 only

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Question:** Determine for what \( x \)-value(s) \( f(x) \) shown below is discontinuous. \[ f(x) = \begin{cases} \frac{x^2 + 2x}{x^2 - 4} & x < 1 \\ -1 & x = 1 \\ 3 - 4e^{x-1} & x > 1 \end{cases} \] **Options:** - \( x = -2 \) and \( x = 2 \) only - \( x = 2 \) only - \( x = 1 \) only - \( x = -2, 1, \text{ and } 2 \) only - \( x = -2 \) only
Transcribed Image Text:**Question:** Determine for what \( x \)-value(s) \( f(x) \) shown below is discontinuous. \[ f(x) = \begin{cases} \frac{x^2 + 2x}{x^2 - 4} & x < 1 \\ -1 & x = 1 \\ 3 - 4e^{x-1} & x > 1 \end{cases} \] **Options:** - \( x = -2 \) and \( x = 2 \) only - \( x = 2 \) only - \( x = 1 \) only - \( x = -2, 1, \text{ and } 2 \) only - \( x = -2 \) only
Expert Solution
Step 1

Check for discontinuous point, we discuss in three case. 

  • When x=1
  • When x <1
  • When x >1
steps

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