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

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter11: Rational And Irrational Numbers
Section: Chapter Questions
Problem 23CLR
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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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