MA 3972-MA-Book May 8, 2018 13:46Review of Precalculus 59(d) (f−g)(1)
(e) (fg)(2)
(f)(
f
g)
(0)(g)(
f
g)
(5)(h)
(
g
f)
(4)Solutions
(a) (f◦g)(x)= f(g(x))= f(x−5)=(x−5)^2 − 4 =x^2 − 10 x+21.
Thus (f◦g)(−1)=(−1)^2 −10(−1)+ 21 = 1 + 10 + 21 =32.
Or (f ◦g)(−1)=f(g(−1))= f(−6)=32.
(b) (g◦f)(x)=g(f(x))=g(x^2 −4)=(x^2 −4)− 5 =x^2 −9.
Thus (g◦f)(−1)=(−1)^2 − 9 = 1 − 9 =−8.
(c) (f+g)(x)=(x^2 −4)+(x−5)=x^2 +x−9. Thus (f +g)(−3)=−3.
(d) (f−g)(x)=(x^2 −4)−(x−5)=x^2 −x+1. Thus (f −g)(1)=1.
(e) (fg)(x)=(x^2 −4)(x−5)=x^3 − 5 x^2 − 4 x+20. Thus (fg)(2)=0.
(f)
(
f
g)
(x)=
x^2 − 4
x− 5
,x =5. Thus(
f
g)
(0)=4
5
.
(g) Sinceg(5)=0,x=5isnotin the domain of(
f
g)
and(
f
g)
(5) isundefined.(h)
(
g
f)
(x)=
x− 5
x^2 − 4
,x =2or−2. Thus(
g
f)
(4)=−1
12
.
Example 2
Givenh(x)=√
xandk(x)=√
9 −x^2 :(a) find(
h
k)
(x) and indicate its domain and(b) find
(
k
h)
(x) and indicate its domain.Solutions(a)(
h
k)
(x)=√
x
√
9 −x^2
The domain ofh(x)is[0,∞), and the domain ofk(x)is[−3, 3].
The intersection of the two domains is [0, 3]. However,k(3)=0.
Therefore, the domain of(
h
k)
is [0, 3).Note that√
x
√
9 −x^2is not equivalent to√
x
9 −x^2
outside of the domain [0, 3).(b)
(
k
h)
(x)=√
9 −x^2
√
x
The intersection of the two domains is [0, 3]. However,h(0)=0.
Therefore, the domain of(
k
h)
is (0, 3].