Find f(g(x)) f(x) = cube root of 3x^2-1 , g(x)=3x^(2/3)+1

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (3 x^{\frac{2}{3}} + 1)$ by substituting in the value of $g$ into $f$ .

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

Step 3

Rewrite ${(3 x^{\frac{2}{3}} + 1)}^{2}$ as $(3 x^{\frac{2}{3}} + 1) (3 x^{\frac{2}{3}} + 1)$ .

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

Step 4

Expand $(3 x^{\frac{2}{3}} + 1) (3 x^{\frac{2}{3}} + 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Step 5

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $x^{\frac{2}{3}}$ by $x^{\frac{2}{3}}$ by adding the exponents.

Move $x^{\frac{2}{3}}$ .

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

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

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

Combine the numerators over the common denominator.

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

Add $2$ and $2$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (3 \cdot (3 x^{\frac{4}{3}}) + 3 x^{\frac{2}{3}} \cdot 1 + 1 (3 x^{\frac{2}{3}}) + 1 \cdot 1) - 1}$

Multiply $3$ by $3$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}} + 3 x^{\frac{2}{3}} \cdot 1 + 1 (3 x^{\frac{2}{3}}) + 1 \cdot 1) - 1}$

Multiply $3$ by $1$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}} + 3 x^{\frac{2}{3}} + 1 (3 x^{\frac{2}{3}}) + 1 \cdot 1) - 1}$

Multiply $3 x^{\frac{2}{3}}$ by $1$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}} + 3 x^{\frac{2}{3}} + 3 x^{\frac{2}{3}} + 1 \cdot 1) - 1}$

Multiply $1$ by $1$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}} + 3 x^{\frac{2}{3}} + 3 x^{\frac{2}{3}} + 1) - 1}$

Add $3 x^{\frac{2}{3}}$ and $3 x^{\frac{2}{3}}$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}} + 6 x^{\frac{2}{3}} + 1) - 1}$

Step 6

Apply the distributive property.

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{3 (9 x^{\frac{4}{3}}) + 3 (6 x^{\frac{2}{3}}) + 3 \cdot 1 - 1}$

Step 7

Simplify.

Multiply $9$ by $3$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{27 x^{\frac{4}{3}} + 3 (6 x^{\frac{2}{3}}) + 3 \cdot 1 - 1}$

Multiply $6$ by $3$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{27 x^{\frac{4}{3}} + 18 x^{\frac{2}{3}} + 3 \cdot 1 - 1}$

Multiply $3$ by $1$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{27 x^{\frac{4}{3}} + 18 x^{\frac{2}{3}} + 3 - 1}$

Step 8

Subtract $1$ from $3$ .

$f (3 x^{\frac{2}{3}} + 1) = \sqrt[3]{27 x^{\frac{4}{3}} + 18 x^{\frac{2}{3}} + 2}$

Find f(g(x)) f(x) = cube root of 3x^2-1 , g(x)=3x^(2/3)+1

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