Find g(f(x)) f(x)=4x-1 g(x)=3x^3-1

$f (x) = 4 x - 1$ $g (x) = 3 x^{3} - 1$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (4 x - 1)$ by substituting in the value of $f$ into $g$ .

$g (4 x - 1) = 3 {(4 x - 1)}^{3} - 1$

Step 3

Simplify each term.

Use the Binomial Theorem.

$g (4 x - 1) = 3 ({(4 x)}^{3} + 3 {(4 x)}^{2} \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Simplify each term.

Apply the product rule to $4 x$ .

$g (4 x - 1) = 3 (4^{3} x^{3} + 3 {(4 x)}^{2} \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Raise $4$ to the power of $3$ .

$g (4 x - 1) = 3 (64 x^{3} + 3 {(4 x)}^{2} \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Apply the product rule to $4 x$ .

$g (4 x - 1) = 3 (64 x^{3} + 3 (4^{2} x^{2}) \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Raise $4$ to the power of $2$ .

$g (4 x - 1) = 3 (64 x^{3} + 3 (16 x^{2}) \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Multiply $16$ by $3$ .

$g (4 x - 1) = 3 (64 x^{3} + 48 x^{2} \cdot - 1 + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Multiply $- 1$ by $48$ .

$g (4 x - 1) = 3 (64 x^{3} - 48 x^{2} + 3 (4 x) {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Multiply $4$ by $3$ .

$g (4 x - 1) = 3 (64 x^{3} - 48 x^{2} + 12 x {(- 1)}^{2} + {(- 1)}^{3}) - 1$

Raise $- 1$ to the power of $2$ .

$g (4 x - 1) = 3 (64 x^{3} - 48 x^{2} + 12 x \cdot 1 + {(- 1)}^{3}) - 1$

Multiply $12$ by $1$ .

$g (4 x - 1) = 3 (64 x^{3} - 48 x^{2} + 12 x + {(- 1)}^{3}) - 1$

Raise $- 1$ to the power of $3$ .

$g (4 x - 1) = 3 (64 x^{3} - 48 x^{2} + 12 x - 1) - 1$

Apply the distributive property.

$g (4 x - 1) = 3 (64 x^{3}) + 3 (- 48 x^{2}) + 3 (12 x) + 3 \cdot - 1 - 1$

Simplify.

Multiply $64$ by $3$ .

$g (4 x - 1) = 192 x^{3} + 3 (- 48 x^{2}) + 3 (12 x) + 3 \cdot - 1 - 1$

Multiply $- 48$ by $3$ .

$g (4 x - 1) = 192 x^{3} - 144 x^{2} + 3 (12 x) + 3 \cdot - 1 - 1$

Multiply $12$ by $3$ .

$g (4 x - 1) = 192 x^{3} - 144 x^{2} + 36 x + 3 \cdot - 1 - 1$

Multiply $3$ by $- 1$ .

$g (4 x - 1) = 192 x^{3} - 144 x^{2} + 36 x - 3 - 1$

Step 4

Subtract $1$ from $- 3$ .

$g (4 x - 1) = 192 x^{3} - 144 x^{2} + 36 x - 4$

Find g(f(x)) f(x)=4x-1 g(x)=3x^3-1

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