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lim f(x) = 8. x → 2+ As x approaches 2 from the left, f(x) approaches 4. As x approaches 2 from the right, f(x) approaches 8. As x approaches 2, f(x) approaches 4, but f(2)= 8. As x approaches 2, f(x) approaches 8, but f(2)= 4. As x approaches 2 from the right, f(x) approaches 4. As x approaches 2 from the left, f(x) approaches 8. Explain what it means to say that lim f(x) = 4 and x → 2- In this situation is it possible that lim f(x) exists? Explain. X→ 2 Yes, f(x) could have a hole at (2, 4) and be defined such that f(2)= 8. Yes, f(x) could have a hole at (2, 8) and be defined such that f(2)= 4. Yes, if f(x) has a vertical asymptote at x = 2, it can be defined such that_lim_ f(x) = 4, lim f(x) = 8, and lim f(x) exists. X-2- X→2+ No, lim f(x) cannot exist if lim_f(x) # limf(x). X-2- X→2 x→2+

lim f(x) = 8. x → 2+ As x approaches 2 from the left, f(x) approaches 4. As x approaches 2 from the right, f(x) approaches 8. As x approaches 2, f(x) approaches 4, but f(2)= 8. As x approaches 2, f(x) approaches 8, but f(2)= 4. As x approaches 2 from the right, f(x) approaches 4. As x approaches 2 from the left, f(x) approaches 8. Explain what it means to say that lim f(x) = 4 and x → 2- In this situation is it possible that lim f(x) exists? Explain. X→ 2 Yes, f(x) could have a hole at (2, 4) and be defined such that f(2)= 8. Yes, f(x) could have a hole at (2, 8) and be defined such that f(2)= 4. Yes, if f(x) has a vertical asymptote at x = 2, it can be defined such that_lim_ f(x) = 4, lim f(x) = 8, and lim f(x) exists. X-2- X→2+ No, lim f(x) cannot exist if lim_f(x) # limf(x). X-2- X→2 x→2+

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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Explain what it means to say that lim f(x) = 4 and x → 2- lim f(x) = 8. X→ 2 As x approaches 2 from the left, f(x) approaches 4. As x approaches 2 from the right, f(x) approaches 8. As x approaches 2, f(x) approaches 4, but f(2) = 8. As x approaches 2, f(x) approaches 8, but f(2) = 4. As x approaches 2 from the right, f(x) approaches 4. As x approaches 2 from the left, f(x) approaches 8. In this situation is it possible that lim f(x) exists? Explain. X→ 2 No, lim f(x) cannot exist if lim_f(x) ‡ x → 2 X→2- + Yes, f(x) could have a hole at (2, 4) and be defined such that f(2) = 8. Yes, f(x) could have a hole at (2, 8) and be defined such that f(2) = 4. Yes, if f(x) has a vertical asymptote at x = 2, it can be defined such that_lim_f(x) = 4, X→2 lim f(x). x→2+ limf(x) 4, lim f(x) = 8, and lim f(x) exists. X→2+ X→ 2
Transcribed Image Text:Explain what it means to say that lim f(x) = 4 and x → 2- lim f(x) = 8. X→ 2 As x approaches 2 from the left, f(x) approaches 4. As x approaches 2 from the right, f(x) approaches 8. As x approaches 2, f(x) approaches 4, but f(2) = 8. As x approaches 2, f(x) approaches 8, but f(2) = 4. As x approaches 2 from the right, f(x) approaches 4. As x approaches 2 from the left, f(x) approaches 8. In this situation is it possible that lim f(x) exists? Explain. X→ 2 No, lim f(x) cannot exist if lim_f(x) ‡ x → 2 X→2- + Yes, f(x) could have a hole at (2, 4) and be defined such that f(2) = 8. Yes, f(x) could have a hole at (2, 8) and be defined such that f(2) = 4. Yes, if f(x) has a vertical asymptote at x = 2, it can be defined such that_lim_f(x) = 4, X→2 lim f(x). x→2+ limf(x) 4, lim f(x) = 8, and lim f(x) exists. X→2+ X→ 2
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. ex = 7 - 6x, (0, 1) and f(1) = . Since The equation e* = 7 - 6x is equivalent to the equation f(x) = e* − 7 + 6x = 0. f(x) is continuous on the interval [0, 1], f(0) there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation e× = 7 − 6x, in the interval (0, 1). - ---Select--- < 0 <---Select--- ✓ I = I
Transcribed Image Text:Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. ex = 7 - 6x, (0, 1) and f(1) = . Since The equation e* = 7 - 6x is equivalent to the equation f(x) = e* − 7 + 6x = 0. f(x) is continuous on the interval [0, 1], f(0) there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation e× = 7 − 6x, in the interval (0, 1). - ---Select--- < 0 <---Select--- ✓ I = I
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