Graph f(x)=| cube root of x+2x|+12

$f (x) = | \sqrt[3]{x} + 2 x | + 12$

Step 1

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 2

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

Substitute the $x$ value $- 2$ into $f (x) = | \sqrt[3]{x} + 2 x | + 12$ . In this case, the point is $(- 2, \sqrt[3]{2} + 16)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = | \sqrt[3]{- 2} + 2 (- 2) | + 12$

Simplify the result.

Remove parentheses.

$f (- 2) = | \sqrt[3]{- 2} + 2 (- 2) | + 12$

Simplify each term.

Rewrite $- 2$ as ${(- 1)}^{3} \cdot 2$ .

Rewrite $- 2$ as $- 1 (2)$ .

$f (- 2) = | \sqrt[3]{- 1 \cdot 2} + 2 (- 2) | + 12$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$f (- 2) = | \sqrt[3]{{(- 1)}^{3} \cdot 2} + 2 (- 2) | + 12$

Pull terms out from under the radical.

$f (- 2) = | - 1 \sqrt[3]{2} + 2 (- 2) | + 12$

Rewrite $- 1 \sqrt[3]{2}$ as $- \sqrt[3]{2}$ .

$f (- 2) = | - \sqrt[3]{2} + 2 (- 2) | + 12$

Multiply $2$ by $- 2$ .

$f (- 2) = | - \sqrt[3]{2} - 4 | + 12$

$- \sqrt[3]{2} - 4$ is approximately $- 5.25992104$ which is negative so negate $- \sqrt[3]{2} - 4$ and remove the absolute value

$f (- 2) = - (- \sqrt[3]{2} - 4) + 12$

Apply the distributive property.

$f (- 2) = \sqrt[3]{2} + 4 + 12$

Multiply $- - \sqrt[3]{2}$ .

Multiply $- 1$ by $- 1$ .

$f (- 2) = 1 \sqrt[3]{2} + 4 + 12$

Multiply $\sqrt[3]{2}$ by $1$ .

$f (- 2) = \sqrt[3]{2} + 4 + 12$

Multiply $- 1$ by $- 4$ .

$f (- 2) = \sqrt[3]{2} + 4 + 12$

Add $4$ and $12$ .

$f (- 2) = \sqrt[3]{2} + 16$

The final answer is $\sqrt[3]{2} + 16$ .

$y = \sqrt[3]{2} + 16$

Substitute the $x$ value $- 1$ into $f (x) = | \sqrt[3]{x} + 2 x | + 12$ . In this case, the point is $(- 1, 15)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = | \sqrt[3]{- 1} + 2 (- 1) | + 12$

Simplify the result.

Remove parentheses.

$f (- 1) = | \sqrt[3]{- 1} + 2 (- 1) | + 12$

Simplify each term.

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$f (- 1) = | \sqrt[3]{{(- 1)}^{3}} + 2 (- 1) | + 12$

Pull terms out from under the radical, assuming real numbers.

$f (- 1) = | - 1 + 2 (- 1) | + 12$

Multiply $2$ by $- 1$ .

$f (- 1) = | - 1 - 2 | + 12$

Subtract $2$ from $- 1$ .

$f (- 1) = | - 3 | + 12$

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f (- 1) = 3 + 12$

Add $3$ and $12$ .

$f (- 1) = 15$

The final answer is $15$ .

$y = 15$

Substitute the $x$ value $0$ into $f (x) = | \sqrt[3]{x} + 2 x | + 12$ . In this case, the point is $(0, 12)$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = | \sqrt[3]{0} + 2 (0) | + 12$

Simplify the result.

Remove parentheses.

$f (0) = | \sqrt[3]{0} + 2 (0) | + 12$

Simplify each term.

Rewrite $0$ as $0^{3}$ .

$f (0) = | \sqrt[3]{0^{3}} + 2 (0) | + 12$

Pull terms out from under the radical, assuming real numbers.

$f (0) = | 0 + 2 (0) | + 12$

Multiply $2$ by $0$ .

$f (0) = | 0 + 0 | + 12$

Add $0$ and $0$ .

$f (0) = | 0 | + 12$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$f (0) = 0 + 12$

Add $0$ and $12$ .

$f (0) = 12$

The final answer is $12$ .

$y = 12$

Substitute the $x$ value $1$ into $f (x) = | \sqrt[3]{x} + 2 x | + 12$ . In this case, the point is $(1, 15)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = | \sqrt[3]{1} + 2 (1) | + 12$

Simplify the result.

Remove parentheses.

$f (1) = | \sqrt[3]{1} + 2 (1) | + 12$

Simplify each term.

Any root of $1$ is $1$ .

$f (1) = | 1 + 2 (1) | + 12$

Multiply $2$ by $1$ .

$f (1) = | 1 + 2 | + 12$

Add $1$ and $2$ .

$f (1) = | 3 | + 12$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$f (1) = 3 + 12$

Add $3$ and $12$ .

$f (1) = 15$

The final answer is $15$ .

$y = 15$

The absolute value can be graphed using the points around the vertex $(- 2, 17.26), (- 1, 15), (0, 12), (1, 15)$

$\begin{array}{c} x & y \\ - 2 & 17.26 \\ - 1 & 15 \\ 0 & 12 \\ 1 & 15 \end{array}$

Step 3

Graph f(x)=| cube root of x+2x|+12

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