College Algebra, 2013a

428 Exponential and Logarithmic Functions
(
f −1 ◦ f ) (x) = f −1 (f(x))
= f −1 ( 2 x−1 − 3 )
=
=
([
log 2 2 x−1 − 3 ] +3 ) +1
(
log 2 2
x−1 ) +1
= (x − 1) + 1 Since log 2 (2 u )=u for all real numbers u
= x ̌
For all real numbers x>−3, we have 11
(
f ◦ f
−1 ) (x) = f ( f −1 (x) )
= f (log 2 (x +3)+1)
= 2 (log 2 (x+3)+1)−1 − 3
= 2 log 2 (x+3) − 3
= (x +3)− 3 Since 2 log 2 (u) = u for all real numbers u>0
= x ̌
5. Last, but certainly not least, we graph y = f(x) andy = f −1 (x) on the same set of axes and
see the symmetry about the line y = x.
8
7
6
5
4
3
2
1
y
−3 −2 −1 1 2 3 4 5 6 7 8
−1
−2
x
y = f(x) =2 x−1 − 3
y = f −1 (x) =log 2 (x +3)+1
11 Pay attention - can you spot in which step below we need x>−3?