Precalculus Examples

If $g (x) = x^{2} + 6$ and $f (x) = 2 x - 4$ ; find $g (f (x))$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (2 x - 4)$ by substituting in the value of $f$ into $g$ .

$g (2 x - 4) = {(2 x - 4)}^{2} + 6$

Step 3

Simplify each term.

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Step 3.1

Rewrite ${(2 x - 4)}^{2}$ as $(2 x - 4) (2 x - 4)$ .

$g (2 x - 4) = (2 x - 4) (2 x - 4) + 6$

Step 3.2

Expand $(2 x - 4) (2 x - 4)$ using the FOIL Method.

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Step 3.2.1

Apply the distributive property.

$g (2 x - 4) = 2 x (2 x - 4) - 4 (2 x - 4) + 6$

Step 3.2.2

Apply the distributive property.

$g (2 x - 4) = 2 x (2 x) + 2 x \cdot - 4 - 4 (2 x - 4) + 6$

Step 3.2.3

Apply the distributive property.

$g (2 x - 4) = 2 x (2 x) + 2 x \cdot - 4 - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3

Simplify and combine like terms.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Rewrite using the commutative property of multiplication.

$g (2 x - 4) = 2 \cdot (2 x \cdot x) + 2 x \cdot - 4 - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3.1.2

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

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Step 3.3.1.2.1

Move $x$ .

$g (2 x - 4) = 2 \cdot (2 (x \cdot x)) + 2 x \cdot - 4 - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3.1.2.2

Multiply $x$ by $x$ .

$g (2 x - 4) = 2 \cdot (2 x^{2}) + 2 x \cdot - 4 - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3.1.3

Multiply $2$ by $2$ .

$g (2 x - 4) = 4 x^{2} + 2 x \cdot - 4 - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3.1.4

Multiply $- 4$ by $2$ .

$g (2 x - 4) = 4 x^{2} - 8 x - 4 (2 x) - 4 \cdot - 4 + 6$

Step 3.3.1.5

Multiply $2$ by $- 4$ .

$g (2 x - 4) = 4 x^{2} - 8 x - 8 x - 4 \cdot - 4 + 6$

Step 3.3.1.6

Multiply $- 4$ by $- 4$ .

$g (2 x - 4) = 4 x^{2} - 8 x - 8 x + 16 + 6$

Step 3.3.2

Subtract $8 x$ from $- 8 x$ .

$g (2 x - 4) = 4 x^{2} - 16 x + 16 + 6$

Step 4

Add $16$ and $6$ .

$g (2 x - 4) = 4 x^{2} - 16 x + 22$