Graph f(x)=|-x|^(1/3)

$f (x) = {| - x |}^{\frac{1}{3}}$

Find the absolute value vertex. In this case, the vertex for $y = {| x |}^{\frac{1}{3}}$ is $(0, 0)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x$ equal to $0$ . In this case, $x = 0$ .

$x = 0$

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

$y = {| 0 |}^{\frac{1}{3}}$

Simplify ${| 0 |}^{\frac{1}{3}}$ .

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

$y = 0^{\frac{1}{3}}$

Simplify the expression.

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

$y = {(0^{3})}^{\frac{1}{3}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 0^{3 (\frac{1}{3})}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = 0^{3 (\frac{1}{3})}$

Rewrite the expression.

$y = 0$

Evaluate the exponent.

$y = 0$

The absolute value vertex is $(0, 0)$ .

$(0, 0)$

Find the domain for $f (x) = {| x |}^{\frac{1}{3}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the absolute value function.

Convert expressions with fractional exponents to radicals.

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\sqrt[3]{{| x |}^{1}}$

Anything raised to $1$ is the base itself.

$\sqrt[3]{| x |}$

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:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

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 $f (x) = {| x |}^{\frac{1}{3}}$ . In this case, the point is $(- 2, 2^{\frac{1}{3}})$ .

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

$f (- 2) = {| - 2 |}^{\frac{1}{3}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (- 2) = 2^{\frac{1}{3}}$

The final answer is $2^{\frac{1}{3}}$ .

$y = 2^{\frac{1}{3}}$

Substitute the $x$ value $- 1$ into $f (x) = {| x |}^{\frac{1}{3}}$ . In this case, the point is $(- 1, 1)$ .

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

$f (- 1) = {| - 1 |}^{\frac{1}{3}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (- 1) = 1^{\frac{1}{3}}$

One to any power is one.

$f (- 1) = 1$

The final answer is $1$ .

$y = 1$

Substitute the $x$ value $2$ into $f (x) = {| x |}^{\frac{1}{3}}$ . In this case, the point is $(2, 2^{\frac{1}{3}})$ .

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

$f (2) = {| 2 |}^{\frac{1}{3}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (2) = 2^{\frac{1}{3}}$

The final answer is $2^{\frac{1}{3}}$ .

$y = 2^{\frac{1}{3}}$

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, 1.26), (- 1, 1), (1, 1), (2, 1.26)$

$\begin{array}{c} x & y \\ - 2 & 1.26 \\ - 1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 1.26 \end{array}$

Graph f(x)=|-x|^(1/3)

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