Pre-Algebra Examples

$f (x) = \sqrt{12 - 3 x^{2}}$

Step 1

Find the domain for $y = \sqrt{12 - 3 x^{2}}$ 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 $\sqrt{3 (2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$3 (2 + x) (2 - x) \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 + x = 0$

$2 - x = 0$

Step 1.2.2

Set $2 + x$ equal to $0$ and solve for $x$ .

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Step 1.2.2.1

Set $2 + x$ equal to $0$ .

$2 + x = 0$

Step 1.2.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.2.3

Set $2 - x$ equal to $0$ and solve for $x$ .

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Step 1.2.3.1

Set $2 - x$ equal to $0$ .

$2 - x = 0$

Step 1.2.3.2

Solve $2 - x = 0$ for $x$ .

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Step 1.2.3.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 1.2.3.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 1.2.3.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 1.2.3.2.2.2

Simplify the left side.

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Step 1.2.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 1.2.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 1.2.3.2.2.3

Simplify the right side.

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Step 1.2.3.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 1.2.4

The final solution is all the values that make $3 (2 + x) (2 - x) \geq 0$ true.

$x = - 2, 2$

Step 1.2.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$x > 2$

Step 1.2.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.6.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 1.2.6.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 1.2.6.1.2

Replace $x$ with $- 4$ in the original inequality.

$3 (2 - 4) (2 - (- 4)) \geq 0$

Step 1.2.6.1.3

The left side $- 36$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.2

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

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Step 1.2.6.2.1

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.6.2.2

Replace $x$ with $0$ in the original inequality.

$3 (2 + 0) (2 - (0)) \geq 0$

Step 1.2.6.2.3

The left side $12$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.6.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 1.2.6.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 1.2.6.3.2

Replace $x$ with $4$ in the original inequality.

$3 (2 + 4) (2 - (4)) \geq 0$

Step 1.2.6.3.3

The left side $- 36$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 2$ False

$- 2 < x < 2$ True

$x > 2$ False

$x < - 2$ False

$- 2 < x < 2$ True

$x > 2$ False

Step 1.2.7

The solution consists of all of the true intervals.

$- 2 \leq x \leq 2$

Step 1.3

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

Interval Notation:

$[- 2, 2]$

Set-Builder Notation:

${x | - 2 \leq x \leq 2}$

Interval Notation:

$[- 2, 2]$

Set-Builder Notation:

${x | - 2 \leq x \leq 2}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{3 (2 + x) (2 - x)}$ .

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Step 2.1

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

$f (- 2) = \sqrt{3 (2 - 2) (2 - (- 2))}$

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

$f (- 2) = \sqrt{3 (2 - 2) (2 - (- 2))}$

Step 2.2.2

Multiply $- 1$ by $- 2$ .

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

Step 2.2.3

Subtract $2$ from $2$ .

$f (- 2) = \sqrt{3 \cdot (0 (2 + 2))}$

Step 2.2.4

Combine exponents.

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Step 2.2.4.1

Multiply $3$ by $0$ .

$f (- 2) = \sqrt{0 (2 + 2)}$

Step 2.2.4.2

Multiply $0$ by $2 + 2$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

$f (- 2) = 0$

Step 2.2.7

The final answer is $0$ .

$0$

Step 2.3

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

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

Step 2.4

Simplify the result.

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Step 2.4.1

Remove parentheses.

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

Step 2.4.2

Multiply $- 1$ by $2$ .

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

Step 2.4.3

Add $2$ and $2$ .

$f (2) = \sqrt{3 \cdot (4 (2 - 2))}$

Step 2.4.4

Multiply $3$ by $4$ .

$f (2) = \sqrt{12 (2 - 2)}$

Step 2.4.5

Subtract $2$ from $2$ .

$f (2) = \sqrt{12 \cdot 0}$

Step 2.4.6

Multiply $12$ by $0$ .

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

Step 2.4.7

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

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

Step 2.4.8

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

$f (2) = 0$

Step 2.4.9

The final answer is $0$ .

$0$

Step 3

The end points are $(- 2, 0), (2, 0)$ .

$(- 2, 0), (2, 0)$

Step 4

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

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

Step 5