The function f (x) = ! takes the value of -1 at x = -1 and 1 at x = 1. The number is between -1 and 1. By the intermediate value theorem there must be an x between -1 and 1 for which f(x) = }. | Show that there is only one value of x for which f(x) = ;, check whether it is between -1 and 1, and then address the flaw in the preceding argument.

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter2: Functions
Section2.7: Combining Functions
Problem 5E: Let f and g be functions. (a) The function (f+g)(x) is defined for all values of x that are in the...
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The function f(x) = ! takes the value of –1 at x = -1 and 1 at x = 1.
The number ; is between -1 and 1. By the intermediate value theorem
there must be an x between –1 and 1 for which f(x) = ;.
1
Show that there is only one value of x for which f(x) = ;, check whether it
is between –1 and 1, and then address the flaw in the preceding argument.
Transcribed Image Text:The function f(x) = ! takes the value of –1 at x = -1 and 1 at x = 1. The number ; is between -1 and 1. By the intermediate value theorem there must be an x between –1 and 1 for which f(x) = ;. 1 Show that there is only one value of x for which f(x) = ;, check whether it is between –1 and 1, and then address the flaw in the preceding argument.
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