Suppose that the functions p and q are defined as follows. p(x)=2x+1
g(x)=x²-1
9
Find the following.
(a op) (2) = 0
Х
(pog)(2) =

(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

(-5,-28)

i graphed to on desmos.