Find f(g(x)) f(x)=9x^2+1 g(x) = square root of 2x^3

$f (x) = 9 x^{2} + 1$ $g (x) = \sqrt{2 x^{3}}$

Set up the composite result function.

$f (g (x))$

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

$f (\sqrt{2 x^{3}}) = 9 {(\sqrt{2 x^{3}})}^{2} + 1$

Simplify each term.

Rewrite $2 x^{3}$ as $x^{2} \cdot (2 x)$ .

Factor out $x^{2}$ .

$f (\sqrt{2 x^{3}}) = 9 {\sqrt{2 (x^{2} x)}}^{2} + 1$

Reorder $2$ and $x^{2}$ .

$f (\sqrt{2 x^{3}}) = 9 {\sqrt{x^{2} \cdot (2 x)}}^{2} + 1$

Add parentheses.

$f (\sqrt{2 x^{3}}) = 9 {\sqrt{x^{2} \cdot (2 x)}}^{2} + 1$

Pull terms out from under the radical.

$f (\sqrt{2 x^{3}}) = 9 {(x \sqrt{2 x})}^{2} + 1$

Apply the product rule to $x \sqrt{2 x}$ .

$f (\sqrt{2 x^{3}}) = 9 (x^{2} {\sqrt{2 x}}^{2}) + 1$

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

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

$f (\sqrt{2 x^{3}}) = 9 (x^{2} {({(2 x)}^{\frac{1}{2}})}^{2}) + 1$

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

$f (\sqrt{2 x^{3}}) = 9 (x^{2} {(2 x)}^{\frac{1}{2} \cdot 2}) + 1$

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

$f (\sqrt{2 x^{3}}) = 9 (x^{2} {(2 x)}^{\frac{2}{2}}) + 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\sqrt{2 x^{3}}) = 9 (x^{2} {(2 x)}^{\frac{2}{2}}) + 1$

Rewrite the expression.

$f (\sqrt{2 x^{3}}) = 9 (x^{2} (2 x)) + 1$

Simplify.

$f (\sqrt{2 x^{3}}) = 9 (x^{2} (2 x)) + 1$

Multiply $x^{2}$ by $x$ by adding the exponents.

Move $x$ .

$f (\sqrt{2 x^{3}}) = 9 (x \cdot x^{2} \cdot 2) + 1$

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

$f (\sqrt{2 x^{3}}) = 9 (x \cdot x^{2} \cdot 2) + 1$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\sqrt{2 x^{3}}) = 9 (x^{1 + 2} \cdot 2) + 1$

Add $1$ and $2$ .

$f (\sqrt{2 x^{3}}) = 9 (x^{3} \cdot 2) + 1$

Move $2$ to the left of $x^{3}$ .

$f (\sqrt{2 x^{3}}) = 9 (2 \cdot x^{3}) + 1$

Multiply $2$ by $9$ .

$f (\sqrt{2 x^{3}}) = 18 x^{3} + 1$