College Algebra, 2013a

5.1 Function Composition 363
(+) ” (−) 0 (+)
−1 − 3 5
Our domain is (−∞, −1) ∪ [ − 3 5 , ∞) .
7. We find (h ◦ g)(x) by finding h(g(x)).
ˆ inside out: We insert the expression g(x) intoh first to get
(h ◦ g)(x) = h(g(x)) =h ( 2 − √ x +3 )
2 ( 2 − √ x +3 )
= ( √ )
2 − x +3 +1
= 4 − 2√ x +3
3 − √ x +3
ˆ outside in: Weusetheformulaforh(x) firsttoget
(h ◦ g)(x) = h(g(x)) = 2(g(x))
(g(x)) + 1
2 ( 2 − √ x +3 )
= ( √ )
2 − x +3 +1
= 4 − 2√ x +3
3 − √ x +3
To find the domain of h ◦ g, we look to the step before any simplification:
(h ◦ g)(x) = 2 ( 2 − √ x +3 )
(
2 −
√ x +3
)
+1
To keep the square root happy, we require x+3 ≥ 0orx ≥−3. Setting the denominator equal
to zero gives ( 2 − √ x +3 ) +1=0 or √ x + 3 = 3. Squaring both sides gives us x +3=9, or
x =6. Sincex = 6 checks in the original equation, ( 2 − √ x +3 ) +1=0,weknowx =6is
the only zero of the denominator. Hence, the domain of h ◦ g is [−3, 6) ∪ (6, ∞).
8. To find (h ◦ h)(x), we substitute the function h into itself, h(h(x)).
ˆ inside out: We insert the expression h(x) intoh to get
( ) 2x
(h ◦ h)(x) = h(h(x)) = h
x +1