Find p(q(x)) p(x)=x^2+6 q(x) = square root of x+9

$p (x) = x^{2} + 6$ $q (x) = \sqrt{x + 9}$

Step 1

Set up the composite result function.

$p (q (x))$

Step 2

Evaluate $p (\sqrt{x + 9})$ by substituting in the value of $q$ into $p$ .

$p (\sqrt{x + 9}) = {(\sqrt{x + 9})}^{2} + 6$

Step 3

Rewrite ${\sqrt{x + 9}}^{2}$ as $x + 9$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 9}$ as ${(x + 9)}^{\frac{1}{2}}$ .

$p (\sqrt{x + 9}) = {({(x + 9)}^{\frac{1}{2}})}^{2} + 6$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$p (\sqrt{x + 9}) = {(x + 9)}^{\frac{1}{2} \cdot 2} + 6$

Combine $\frac{1}{2}$ and $2$ .

$p (\sqrt{x + 9}) = {(x + 9)}^{\frac{2}{2}} + 6$

Cancel the common factor of $2$ .

Cancel the common factor.

$p (\sqrt{x + 9}) = {(x + 9)}^{\frac{2}{2}} + 6$

Rewrite the expression.

$p (\sqrt{x + 9}) = (x + 9) + 6$

Simplify.

$p (\sqrt{x + 9}) = x + 9 + 6$

Step 4

Add $9$ and $6$ .

$p (\sqrt{x + 9}) = x + 15$