Mathematics: Suppose that the functions p and q are defined as follows.

Suppose that the functions p and q are defined

kylucienne

29.01.2022

The answer is " $\bold{(q \circ p)(2)= 3}\ and \ \bold{(p \circ q)(2)=7}$ " and the further explanation can be defined as follows;

Given:

$\to \bold{p(x)=x^2+3}\\\\\to \bold{q(x)=\sqrt{x+2}}$

Find:

$\bold{(q \circ p)(2)=?}\\\\\bold{(p \circ q)(2)=?}$

Solve the value for $\bold{(q \circ p)(2)}\\\\$ :

$\to \bold{(q \circ p)(2)= q \circ p(2) =q(p(2))}\\\\$

$\therefore\\\\ \to \bold{p(2)=2^2+3= 4+3=7}\\\\\ \because \\\\ \to \bold{q(p(2))=\sqrt{7+2}=\sqrt{9}=3}$

Solve the value for $\bold{(p \circ q)(2)}\\\\$ :

$\to \bold{(p \circ q)(2)= p \circ q(2)= p (q(2))}\\\\$

$\therefore\\\\ \to \bold{q(2)=\sqrt{2+2}=\sqrt{4}=2}\\\\\ \because \\\\ \to \bold{p(q(2))=2^2+3= 4+3=7}$

Therefore the final answer of " $\bold{(q \circ p)(2)= 3}\ and \ \bold{(p \circ q)(2)=7}$ "

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lcyaire

29.01.2022

Step-by-step explanation:

Hello,

qop(2)=q(p(2))

p(2) = 4+3=7

$q(7) = \sqrt{7+2}=\sqrt{9}=3$

qop(2)=3

and poq(2)=p(q(2))

$q(2)=\sqrt{2+2} = \sqrt{4}=2$

p(2) = 7

so poq(2)=7

thanks

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