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

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\sqrt{x - 2})$ by substituting in the value of $g$ into $f$ .

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

Step 3

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Step 4

Add $- 2$ and $3$ .

$f (\sqrt{x - 2}) = x + 1$