Precalculus Examples

If $f (x) = x^{2} + 1$ and $g (x) = \sqrt{2 x + 3}$ ; find $((f \circ g)) (x)$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\sqrt{2 x + 3})$ by substituting in the value of $g$ into $f$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Cancel the common factor of $2$ .

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Step 3.4.1

Cancel the common factor.

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

Step 3.4.2

Rewrite the expression.

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

Step 3.5

Simplify.

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

Step 4

Add $3$ and $1$ .

$f (\sqrt{2 x + 3}) = 2 x + 4$