Algebra Examples

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

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.

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Step 3.1

Use the Binomial Theorem.

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

Step 3.2

Simplify each term.

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Step 3.2.1

Apply the product rule to $4 x$ .

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

Step 3.2.2

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

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

Step 3.2.3

Apply the product rule to $4 x$ .

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

Step 3.2.4

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

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

Step 3.2.5

Multiply $16$ by $3$ .

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

Step 3.2.6

Multiply $- 1$ by $48$ .

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

Step 3.2.7

Multiply $4$ by $3$ .

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

Step 3.2.8

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

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

Step 3.2.9

Multiply $12$ by $1$ .

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

Step 3.2.10

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

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

Step 3.3

Apply the distributive property.

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

Step 3.4

Simplify.

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Step 3.4.1

Multiply $64$ by $3$ .

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

Step 3.4.2

Multiply $- 48$ by $3$ .

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

Step 3.4.3

Multiply $12$ by $3$ .

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

Step 3.4.4

Multiply $3$ by $- 1$ .

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

Step 4

Subtract $1$ from $- 3$ .

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