Guided Example #2 - Composition of Functions
Let (x)=√1-2x , and h(x)=x2+1. Find (h∘f)(-4)
a. Does not exist
b. 10
c. -4
d. 
√7
The correct answer is B. To see why, view each of the steps below.
By definition, %255Cleft(-4%255Cright)%253Dh%255Cleft(f%255Cleft(-4%255Cright)%255Cright) "\left(h\circ f\right)\left(-4\right)=h\left(f\left(-4\right)\right)")
(h∘f)(−4)=h(f(−4)). But %253D%255Csqrt%257B1-2(-4)%257D%253D%255Csqrt%257B1%252B8%257D%253D%255Csqrt9%253D3 "f(-4)=\sqrt{1-2(-4)}=\sqrt{1+8}=\sqrt9=3")
f(−4)=√1−2(−4)=√1+8=√9=3.
Therefore, )%253Dh(3)%253D3%255E2%252B1%253D10 "h(f(-4))=h(3)=3^2+1=10")
h(f(−4))=h(3)=32+1=10.
By definition, (h∘f)(−4)=h(f(−4)). But f(−4)=√1−2(−4)=√1+8=√9=3.
Therefore, h(f(−4))=h(3)=32+1=10.