The answer is "[tex]\bold{(q \circ p)(2)= 3}\ and \ \bold{(p \circ q)(2)=7}[/tex]" and the further explanation can be defined as follows;
Given:
[tex]\to \bold{p(x)=x^2+3}\\\\\to \bold{q(x)=\sqrt{x+2}}[/tex]
Find:
[tex]\bold{(q \circ p)(2)=?}\\\\\bold{(p \circ q)(2)=?}[/tex]
Solve the value for [tex]\bold{(q \circ p)(2)}\\\\[/tex]:
[tex]\to \bold{(q \circ p)(2)= q \circ p(2) =q(p(2))}\\\\[/tex]
[tex]\therefore\\\\ \to \bold{p(2)=2^2+3= 4+3=7}\\\\\ \because \\\\ \to \bold{q(p(2))=\sqrt{7+2}=\sqrt{9}=3}[/tex]
Solve the value for [tex]\bold{(p \circ q)(2)}\\\\[/tex]:
[tex]\to \bold{(p \circ q)(2)= p \circ q(2)= p (q(2))}\\\\[/tex]
[tex]\therefore\\\\ \to \bold{q(2)=\sqrt{2+2}=\sqrt{4}=2}\\\\\ \because \\\\ \to \bold{p(q(2))=2^2+3= 4+3=7}[/tex]
Therefore the final answer of "[tex]\bold{(q \circ p)(2)= 3}\ and \ \bold{(p \circ q)(2)=7}[/tex]"
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