Unformatted text preview: tions
y
6
5 y = j (x) y=x 4
3
2
1 1 2 3 4 5 6 x −1 y = j −1 (x) 2. We graph y = k (x) = √ x + 2 − 1 using what we learned in Section 1.8 and see k is onetoone.
y
2
1 −2 −1 1 2 x −1
−2 y = k (x) We now try to ﬁnd k −1 .
y
y
x
x+1
(x + 1)2
2 + 2x + 1
x
y =
=
=
=
=
=
= k (x)
√
x+2−1
√
y + 2 − 1 switch x and y
√
y+2
√
2
y+2
y+2
x2 + 2 x − 1 We have k −1 (x) = x2 + 2x − 1. Based on our experience, we know something isn’t quite right.
We determined k −1 is a quadratic function, and we have seen several times in this section
that these are not onetoone unless their domains are suitably restricted. Theorem 5.2 tells
us that the domain of k −1 is the range of k . From the graph of k , we see that the range
is [−1, ∞), which means we restrict the domain of k −1 to x ≥ −1. We now check that this
works in our compositions. 5.2 Inverse Functions k −1 ◦ k (x) =
=
=
=
=
= 307 k −1 (k (x))
√
x + 2 − 1 , x ≥ −2
k −1
√
√
2
x+2−1 +2 x+2−1 −1
√
√
√
2
x+2 −2 x+2+1+2 x+2−2−1
x+2−2
x and
k ◦ k −1 (x) =
=
=
=
=
=
= k x2 + 2x − 1 x ≥ −1
2
√ (x + 2x − 1) + 2 − 1
2 + 2x + 1 − 1
x
(x + 1)2 − 1
x + 1 − 1
x+1−1
since x ≥ −1
x Graphically, everything checks out as well, provided that we remember the domain re...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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