Find q(r(x)) q(x)=x^2+6 r(x) = square root of x+1

$q (x) = x^{2} + 6$ $r (x) = \sqrt{x + 1}$

Step 1

Set up the composite result function.

$q (r (x))$

Step 2

Evaluate $q (\sqrt{x + 1})$ by substituting in the value of $r$ into $q$ .

$q (\sqrt{x + 1}) = {(\sqrt{x + 1})}^{2} + 6$

Step 3

Rewrite ${\sqrt{x + 1}}^{2}$ as $x + 1$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$q (\sqrt{x + 1}) = (x + 1) + 6$

Simplify.

$q (\sqrt{x + 1}) = x + 1 + 6$

Step 4

Add $1$ and $6$ .

$q (\sqrt{x + 1}) = x + 7$