Finite Math Examples

$f (x) = \sqrt{x - 3}$ , $g (x) = x^{2} + 12$ , $(g \circ f)$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

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

$g (\sqrt{x - 3}) = {(\sqrt{x - 3})}^{2} + 12$

Step 3

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

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Step 3.1

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

$g (\sqrt{x - 3}) = {({(x - 3)}^{\frac{1}{2}})}^{2} + 12$

Step 3.2

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

$g (\sqrt{x - 3}) = {(x - 3)}^{\frac{1}{2} \cdot 2} + 12$

Step 3.3

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

$g (\sqrt{x - 3}) = {(x - 3)}^{\frac{2}{2}} + 12$

Step 3.4

Cancel the common factor of $2$ .

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Step 3.4.1

Cancel the common factor.

$g (\sqrt{x - 3}) = {(x - 3)}^{\frac{2}{2}} + 12$

Step 3.4.2

Rewrite the expression.

$g (\sqrt{x - 3}) = (x - 3) + 12$

Step 3.5

Simplify.

$g (\sqrt{x - 3}) = x - 3 + 12$

Step 4

Add $- 3$ and $12$ .

$g (\sqrt{x - 3}) = x + 9$