Precalculus Examples

$f (x) = 4 x - 16$ , $g (x) = \frac{x}{4} + 4$

Step 1

Find the sum of the functions.

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Step 1.1

Replace the function designators with the actual functions in $f (x) + g (x)$ .

$4 x - 16 + \frac{x}{4} + 4$

Step 1.2

Simplify.

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Step 1.2.1

To write $4 x$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$4 x \cdot \frac{4}{4} + \frac{x}{4} - 16 + 4$

Step 1.2.2

Combine $4 x$ and $\frac{4}{4}$ .

$\frac{4 x \cdot 4}{4} + \frac{x}{4} - 16 + 4$

Step 1.2.3

Combine the numerators over the common denominator.

$\frac{4 x \cdot 4 + x}{4} - 16 + 4$

Step 1.2.4

Find the common denominator.

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Step 1.2.4.1

Write $- 16$ as a fraction with denominator $1$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16}{1} + 4$

Step 1.2.4.2

Multiply $\frac{- 16}{1}$ by $\frac{4}{4}$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16}{1} \cdot \frac{4}{4} + 4$

Step 1.2.4.3

Multiply $\frac{- 16}{1}$ by $\frac{4}{4}$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16 \cdot 4}{4} + 4$

Step 1.2.4.4

Write $4$ as a fraction with denominator $1$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16 \cdot 4}{4} + \frac{4}{1}$

Step 1.2.4.5

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16 \cdot 4}{4} + \frac{4}{1} \cdot \frac{4}{4}$

Step 1.2.4.6

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$\frac{4 x \cdot 4 + x}{4} + \frac{- 16 \cdot 4}{4} + \frac{4 \cdot 4}{4}$

Step 1.2.5

Combine the numerators over the common denominator.

$\frac{4 x \cdot 4 + x - 16 \cdot 4 + 4 \cdot 4}{4}$

Step 1.2.6

Simplify each term.

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Step 1.2.6.1

Multiply $4$ by $4$ .

$\frac{16 x + x - 16 \cdot 4 + 4 \cdot 4}{4}$

Step 1.2.6.2

Multiply $- 16$ by $4$ .

$\frac{16 x + x - 64 + 4 \cdot 4}{4}$

Step 1.2.6.3

Multiply $4$ by $4$ .

$\frac{16 x + x - 64 + 16}{4}$

Step 1.2.7

Simplify by adding terms.

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Step 1.2.7.1

Add $16 x$ and $x$ .

$\frac{17 x - 64 + 16}{4}$

Step 1.2.7.2

Add $- 64$ and $16$ .

$\frac{17 x - 48}{4}$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3