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...
View
Full Document
 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

Click to edit the document details