Find f(g(x)) f(x)=x^2+10 ; g(x) = square root of x-3

$f (x) = x^{2} + 10$ ; $g (x) = \sqrt{x - 3}$

Set up the composite result function.

$f (g (x))$

Evaluate $f (\sqrt{x - 3})$ by substituting in the value of $g$ into $f$ .

$f (\sqrt{x - 3}) = {(\sqrt{x - 3})}^{2} + 10$

Rewrite ${\sqrt{x - 3}}^{2}$ as $x - 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - 3}$ as ${(x - 3)}^{\frac{1}{2}}$ .

$f (\sqrt{x - 3}) = {({(x - 3)}^{\frac{1}{2}})}^{2} + 10$

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

$f (\sqrt{x - 3}) = {(x - 3)}^{\frac{1}{2} \cdot 2} + 10$

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

$f (\sqrt{x - 3}) = {(x - 3)}^{\frac{2}{2}} + 10$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\sqrt{x - 3}) = {(x - 3)}^{\frac{2}{2}} + 10$

Rewrite the expression.

$f (\sqrt{x - 3}) = (x - 3) + 10$

Simplify.

$f (\sqrt{x - 3}) = x - 3 + 10$

Add $- 3$ and $10$ .

$f (\sqrt{x - 3}) = x + 7$