Stitz-Zeager_College_Algebra_e-book

A f x b f x 1 x2 d f x 3 16x x2 9 4 2 3 c

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Unformatted text preview: g −1 ◦ g (x) = x for all x in the domain of g , that is, for all x = 1. g −1 ◦ g (x) = g −1 (g (x)) = g −1 2x 1−x = 2x 1−x 2x +2 1−x = 2x (1 − x) 1−x · 2x (1 − x) +2 1−x clear denominators 302 Further Topics in Functions = 2x 2x + 2(1 − x) = 2x 2x + 2 − 2x = = 2x 2 x Next, we check g g −1 (x) = x for all x in the range of g . From the graph of g in Example 5.2.1, we have that the range of g is (−∞, −2) ∪ (−2, ∞). This matches the domain we get x from the formula g −1 (x) = x+2 , as it should. g ◦ g −1 (x) = g g −1 (x) x x+2 =g = x x+2 x 1− x+2 = x x+2 x 1− x+2 = 2x (x + 2) − x = 2x 2 2 2 · (x + 2) (x + 2) clear denominators =x Graphing y = g (x) and y = g −1 (x) on the same set of axes is busy, but we can see the symmetric relationship if we thicken the curve for y = g −1 (x). Note that the vertical asymptote x = 1 of the graph of g corresponds to the horizontal asymptote y = 1 of the graph of g −1 , as it should since x and...
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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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