Pre-Algebra Examples

$\sqrt{v^{11}}$

Step 1

Find the domain for $y = \sqrt{v^{11}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 1.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 1.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 2

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = | x^{5} | \sqrt{x}$ .

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Step 2.1

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

$f (0) = | {(0)}^{5} | \sqrt{0}$

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

$f (0) = | {(0)}^{5} | \sqrt{0}$

Step 2.2.2

Raising $0$ to any positive power yields $0$ .

$f (0) = | 0 | \sqrt{0}$

Step 2.2.3

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

$f (0) = 0 \sqrt{0}$

Step 2.2.4

Rewrite $0$ as $0^{2}$ .

$f (0) = 0 \sqrt{0^{2}}$

Step 2.2.5

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

$f (0) = 0 \cdot 0$

Step 2.2.6

Multiply $0$ by $0$ .

$f (0) = 0$

Step 2.2.7

The final answer is $0$ .

$0$

Step 3

The radical expression end point is $(0, 0)$ .

$(0, 0)$

Step 4

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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Step 4.1

Substitute the $x$ value $1$ into $f (x) = | x^{5} | \sqrt{x}$ . In this case, the point is $(1, 1)$ .

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Step 4.1.1

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

$f (1) = | {(1)}^{5} | \sqrt{1}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

$f (1) = | {(1)}^{5} | \sqrt{1}$

Step 4.1.2.2

One to any power is one.

$f (1) = | 1 | \sqrt{1}$

Step 4.1.2.3

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

$f (1) = 1 \sqrt{1}$

Step 4.1.2.4

Multiply $\sqrt{1}$ by $1$ .

$f (1) = \sqrt{1}$

Step 4.1.2.5

Any root of $1$ is $1$ .

$f (1) = 1$

Step 4.1.2.6

The final answer is $1$ .

$y = 1$

Step 4.2

Substitute the $x$ value $2$ into $f (x) = | x^{5} | \sqrt{x}$ . In this case, the point is $(2, 32 \sqrt{2})$ .

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Step 4.2.1

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

$f (2) = | {(2)}^{5} | \sqrt{2}$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Remove parentheses.

$f (2) = | {(2)}^{5} | \sqrt{2}$

Step 4.2.2.2

Raise $2$ to the power of $5$ .

$f (2) = | 32 | \sqrt{2}$

Step 4.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $32$ is $32$ .

$f (2) = 32 \sqrt{2}$

Step 4.2.2.4

The final answer is $32 \sqrt{2}$ .

$y = 32 \sqrt{2}$

Step 4.3

The square root can be graphed using the points around the vertex $(0, 0), (1, 1), (2, 45.25)$

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 2 & 45.25 \end{array}$

Step 5