Certainly! Let's find the compositions [tex]\((p \circ q)(1)\)[/tex] and [tex]\((q \circ p)(1)\)[/tex] step by step.
### Finding [tex]\((p \circ q)(1)\)[/tex]
1. Understand what [tex]\((p \circ q)(1)\)[/tex] means: It represents the function [tex]\(p\)[/tex] applied to the result of the function [tex]\(q\)[/tex] evaluated at [tex]\(1\)[/tex]. In other words, [tex]\(p(q(1))\)[/tex].
2. Evaluate [tex]\(q(1)\)[/tex]:
[tex]\[
q(x) = \sqrt{x + 8}
\][/tex]
Substitute [tex]\(x = 1\)[/tex]:
[tex]\[
q(1) = \sqrt{1 + 8} = \sqrt{9} = 3
\][/tex]
3. Now evaluate [tex]\(p(3)\)[/tex]:
[tex]\[
p(x) = x^2 + 7
\][/tex]
Substitute [tex]\(x = 3\)[/tex]:
[tex]\[
p(3) = 3^2 + 7 = 9 + 7 = 16
\][/tex]
Thus, [tex]\((p \circ q)(1) = 16\)[/tex].
### Finding [tex]\((q \circ p)(1)\)[/tex]
1. Understand what [tex]\((q \circ p)(1)\)[/tex] means: It represents the function [tex]\(q\)[/tex] applied to the result of the function [tex]\(p\)[/tex] evaluated at [tex]\(1\)[/tex]. In other words, [tex]\(q(p(1))\)[/tex].
2. Evaluate [tex]\(p(1)\)[/tex]:
[tex]\[
p(x) = x^2 + 7
\][/tex]
Substitute [tex]\(x = 1\)[/tex]:
[tex]\[
p(1) = 1^2 + 7 = 1 + 7 = 8
\][/tex]
3. Now evaluate [tex]\(q(8)\)[/tex]:
[tex]\[
q(x) = \sqrt{x + 8}
\][/tex]
Substitute [tex]\(x = 8\)[/tex]:
[tex]\[
q(8) = \sqrt{8 + 8} = \sqrt{16} = 4
\][/tex]
Thus, [tex]\((q \circ p)(1) = 4\)[/tex].
### Summary
- [tex]\((p \circ q)(1) = 16\)[/tex]
- [tex]\((q \circ p)(1) = 4\)[/tex]
So the final results are:
[tex]\[
\begin{array}{l}
(p \circ q)(1)= 16 \\
(q \circ p)(1)= 4
\end{array}
\][/tex]
