Stitz-Zeager_College_Algebra_e-book

Solving for z we obtain z 50 x2 900 since z represents

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Unformatted text preview: = −2 + x + =x 1 2 1 2 We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers. (Recall that the domain of f −1 ) is the range of f .) f ◦ f −1 (x) = f (f −1 (x)) 1 − 2f −1 (x) = 5 5 1 − 2 −2x + = 5 1 + 5x − 1 = 5 5x = 5 =x 1 2 To check our answer graphically, we graph y = f (x) and y = f −1 (x) on the same set of axes.5 They appear to be reflections across the line y = x. 5 Note that if you perform your check on a calculator for more sophisticated functions, you’ll need to take advantage of the ‘ZoomSquare’ feature to get the correct geometric perspective. 5.2 Inverse Functions 301 y 2 y=x 1 −4 −3 −2 −1 1 2 3 4 x −1 y = f (x) −2 y = f −1 (x) 2. To find g −1 (x), we start with y = g (x). We note that the domain of g is (−∞, 1) ∪ (1, ∞). y = g (x) 2x y= 1−x 2y x= 1−y x(1 − y ) = 2y switch x and y x − xy = 2y x = xy + 2y x = y (x + 2) y= factor x x+2 x We obtain g −1 (x) = x+2 . To check this analytically, we first check...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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