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The function f is continuous where f(-5) = -1 and f(5) = 6, and g is a function defined by g(x) = 1 - (f(x))². Is there a value c for -5 ≤ c ≤ 5 such that g(c) = 1? Why, or why not? Yes; the function g is continuous Yes; g(-5) < 1 < g(5), so IVT guarantees there is a value c for -5 < c < 5 such that g(c) = 1 O No; 1 is not between g(-5) and g(5), so IVT cannot guarantee there is a value c for -5 < c < 5 such that g(c) = 1 No; the function g is not continuous

The function f is continuous where f(-5) = -1 and f(5) = 6, and g is a function defined by g(x) = 1 - (f(x))². Is there a value c for -5 ≤ c ≤ 5 such that g(c) = 1? Why, or why not? Yes; the function g is continuous Yes; g(-5) < 1 < g(5), so IVT guarantees there is a value c for -5 < c < 5 such that g(c) = 1 O No; 1 is not between g(-5) and g(5), so IVT cannot guarantee there is a value c for -5 < c < 5 such that g(c) = 1 No; the function g is not continuous

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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- The function f is continuous where f(-5) = −1 and f(5) = 6, and g is a function defined by g(x) = 1 − (f(x))². Is there a value c for -5 ≤ c ≤ 5 such that g(c) = 1? Why, or why not? Yes; the function g is continuous Yes; g(-5) < 1 < g(5), so IVT guarantees there is a value c for -5 < c < 5 such that g(c) = 1 No; 1 is not between g(-5) and g(5), so IVT cannot guarantee there is a value c for -5 < c < 5 such that g(c) = 1 No; the function g is not continuous
Transcribed Image Text:- The function f is continuous where f(-5) = −1 and f(5) = 6, and g is a function defined by g(x) = 1 − (f(x))². Is there a value c for -5 ≤ c ≤ 5 such that g(c) = 1? Why, or why not? Yes; the function g is continuous Yes; g(-5) < 1 < g(5), so IVT guarantees there is a value c for -5 < c < 5 such that g(c) = 1 No; 1 is not between g(-5) and g(5), so IVT cannot guarantee there is a value c for -5 < c < 5 such that g(c) = 1 No; the function g is not continuous
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