Precalculus Examples

$f (x) = 2 (x) - 1$ , $g (x) = x^{2}$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (2 (x) - 1)$ by substituting in the value of $f$ into $g$ .

$g (2 (x) - 1) = {(2 (x) - 1)}^{2}$

Step 3

Rewrite ${(2 x - 1)}^{2}$ as $(2 x - 1) (2 x - 1)$ .

$g (2 (x) - 1) = (2 x - 1) (2 x - 1)$

Step 4

Expand $(2 x - 1) (2 x - 1)$ using the FOIL Method.

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Step 4.1

Apply the distributive property.

$g (2 (x) - 1) = 2 x (2 x - 1) - 1 (2 x - 1)$

Step 4.2

Apply the distributive property.

$g (2 (x) - 1) = 2 x (2 x) + 2 x \cdot - 1 - 1 (2 x - 1)$

Step 4.3

Apply the distributive property.

$g (2 (x) - 1) = 2 x (2 x) + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1$

Step 5

Simplify and combine like terms.

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Step 5.1

Simplify each term.

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Step 5.1.1

Rewrite using the commutative property of multiplication.

$g (2 (x) - 1) = 2 \cdot (2 x \cdot x) + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1$

Step 5.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 5.1.2.1

Move $x$ .

$g (2 (x) - 1) = 2 \cdot (2 (x \cdot x)) + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1$

Step 5.1.2.2

Multiply $x$ by $x$ .

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

Step 5.1.3

Multiply $2$ by $2$ .

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

Step 5.1.4

Multiply $- 1$ by $2$ .

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

Step 5.1.5

Multiply $2$ by $- 1$ .

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

Step 5.1.6

Multiply $- 1$ by $- 1$ .

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

Step 5.2

Subtract $2 x$ from $- 2 x$ .

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