Exercise1.6
x
.Find f −1 (x) and identify the domain and range of ƒ -1. As a check,
x−3
show that ƒ(ƒ -1(x)) = ƒ -1(ƒ(x)) = x.
32. f (x) =
(a)Tofind f −1 (x) :
Step1:Solveforxintermsofy.
x
y=
⇒ y( x − 3) = x ⇒ y x − 3y = x ⇒ (y − 1) x = 3y x−3
2
3y
⎛ 3y ⎞
⇒ x =
⇒ x=⎜
y −1
⎝ y − 1 ⎟⎠
Step2:Interchangexandy.
2
⎛ 3x ⎞
y=⎜
= f −1 (x) ⎝ x − 1 ⎟⎠
(b)Tofindthedomainandrangeof f −1 (x) :
Thedomainof f (x) is: x ≥ 0 and x ≠ 9 (since x − 3 ≠ 0 ),i.e. ( 9,∞ ) ∪ [ 0,9 ) ,sothe
rangeof f −1 (x) isalso ( 9,∞ ) ∪ [ 0,9 ) .
If 0 ≤ x < 9 ,then 0 ≤ x < 3 ⇒
x − 3 < 0 .Hence f (x) =
x
≤ 0 .
x−3
x
> 1 .
x−3
Therefore,therangeof f (x) is ( −∞,0 ] ∪ (1,∞ ) ,whichmeansthatthedomainof
If x > 9 ,then x > 0 and x − 3 > 0 ⇒
x > x − 3 > 0 .Hence f (x) =
f −1 (x) isalso ( −∞,0 ] ∪ (1,∞ ) .
(c)Toshowthat ƒ(ƒ -1(x)) = ƒ -1(ƒ(x)) = x:
f ( f −1 (x)) =
⎛ 3x ⎞
⎜⎝
⎟
x − 1⎠
2
2
⎛ 3x ⎞
⎜⎝
⎟ −3
x − 1⎠
.
Onthedomainof f −1 (x) ( −∞,0 ] ∪ (1,∞ ) ,wehave
3x
≥ 0 ,so
x −1
3x
3x
3x
f ( f −1 (x)) = x − 1 =
=
= x .
3x
− 3 3x − 3(x − 1) 3
x −1
2
⎛
x ⎞
2
3
⎜
⎟
⎛
⎞
3 x
x
−
3
−1
And f ( f (x)) = ⎜
⎟ =⎜
⎟⎠ = x .
x
x
−
(
x
−
3)
⎝
⎜
− 1⎟
⎝ x−3 ⎠