<h2>Answer:</h2>
Answer:
(r o q)(-1) = 20
(q o r)(-1) = -11
Step-by-step explanation:
Given
q(x) = -2x + 1q(x)=−2x+1
r(x) = 2x^2 + 2r(x)=2x2+2
Solving (a): (r o q)(-1)
In function:
(r o q)(x) = r(q(x))
So, first we calculate q(-1)
q(x) = -2x + 1q(x)=−2x+1
q(-1) = -2(-1) + 1q(−1)=−2(−1)+1
q(-1) = 2 + 1q(−1)=2+1
q(-1) = 3q(−1)=3
Next, we calculate r(q(-1))
Substitute 3 for q(-1)in r(q(-1))
r(q(-1)) = r(3)
This gives:
r(x) = 2x^2 + 2r(x)=2x2+2
r(3) = 2(3)^2 + 2r(3)=2(3)2+2
r(-1) = 2*9 + 2r(−1)=2∗9+2
r(-1) = 20r(−1)=20
Hence:
(r o q)(-1) = 20
Solving (b): (q o r)(-1)
So, first we calculate r(-1)
r(x) = 2x^2 + 2r(x)=2x2+2
r(-1) = 2(-1)^2 + 2r(−1)=2(−1)2+2
r(-1) = 2*1 + 2r(−1)=2∗1+2
\begin{gathered}r(-1) = 6\\\end{gathered}r(−1)=6
Next, we calculate r(q(-1))
Substitute 6 for r(-1)in q(r(-1))
q(r(-1)) = q(6)
q(x) = -2x + 1q(x)=−2x+1
q(6) = -2(6) + 1q(6)=−2(6)+1
q(6) =- 12 + 1q(6)=−12+1
q(6) = -11q(6)=−11
Hence:
(q o r)(-1) = -11
