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The function f-1 is an inverse function of ƒ if f(ƒ −¹(x)) = f−¹(f(x)). To verify, consider f(x) = 5x f(f−¹(x)) = f−¹(f(x)) f(F-¹(x)) = f (X + 8) 5 Hence, = 5(x + 8) - = x + 8 - 8 || 8 and f-¹(x): ƒ−¹(f(x)) = f−¹( || f(ƒ −¹(x)) = ƒ −¹(f(x)) 8 (5x – 8) + 8 5 = X + 8 5 - 8)

The function f-1 is an inverse function of ƒ if f(ƒ −¹(x)) = f−¹(f(x)). To verify, consider f(x) = 5x f(f−¹(x)) = f−¹(f(x)) f(F-¹(x)) = f (X + 8) 5 Hence, = 5(x + 8) - = x + 8 - 8 || 8 and f-¹(x): ƒ−¹(f(x)) = f−¹( || f(ƒ −¹(x)) = ƒ −¹(f(x)) 8 (5x – 8) + 8 5 = X + 8 5 - 8)

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.7: Inverse Functions
Problem 2SE: Why do we restrict the domain of the function f(x)=x2 to find the function's inverse?
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The function f-1 is an inverse function of f if f(ƒ −¹(x)) = ƒ−¹(f(x)). To verify, consider f(x) = 5x – 8 and ƒ −¹(x) f(f−¹(x)) = f−¹(f(x)) f(f-1(x)) = f (x + 8) 5 Hence, = 5(x + 8). 5 = x + 8 - 8 = ƒ−¹(f(x)) = f−¹( || (5x - 8 8) + 8 5 f(f−¹(x)) = f−¹(f(x)) So the domain of f inverse is not = X + 8 5 - 8) the range of f.
Transcribed Image Text:The function f-1 is an inverse function of f if f(ƒ −¹(x)) = ƒ−¹(f(x)). To verify, consider f(x) = 5x – 8 and ƒ −¹(x) f(f−¹(x)) = f−¹(f(x)) f(f-1(x)) = f (x + 8) 5 Hence, = 5(x + 8). 5 = x + 8 - 8 = ƒ−¹(f(x)) = f−¹( || (5x - 8 8) + 8 5 f(f−¹(x)) = f−¹(f(x)) So the domain of f inverse is not = X + 8 5 - 8) the range of f.
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