Precalculus Examples

$y = x \sqrt{1 - x^{2}}$

Step 1

Find the domain for $y = x \sqrt{1 - x^{2}}$ 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 $\sqrt{(1 + x) (1 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(1 + x) (1 - x) \geq 0$

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 $0$ .

$1 + x = 0$

$1 - x = 0$

Step 1.2.2

Set $1 + x$ equal to $0$ and solve for $x$ .

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Step 1.2.2.1

Set $1 + x$ equal to $0$ .

$1 + x = 0$

Step 1.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.3

Set $1 - x$ equal to $0$ and solve for $x$ .

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Step 1.2.3.1

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Step 1.2.3.2

Solve $1 - x = 0$ for $x$ .

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Step 1.2.3.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.2.3.2.2

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

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Step 1.2.3.2.2.1

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

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

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.

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

Step 1.2.3.2.2.2.2

Divide $x$ by $1$ .

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

Step 1.2.3.2.2.3

Simplify the right side.

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Step 1.2.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 1.2.4

The final solution is all the values that make $(1 + x) (1 - x) \geq 0$ true.

$x = - 1, 1$

Step 1.2.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

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 $x < - 1$ to see if it makes the inequality true.

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Step 1.2.6.1.1

Choose a value on the interval $x < - 1$ 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.

$(1 - 4) (1 - (- 4)) \geq 0$

Step 1.2.6.1.3

The left side $- 15$ 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 $- 1 < x < 1$ to see if it makes the inequality true.

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Step 1.2.6.2.1

Choose a value on the interval $- 1 < x < 1$ 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.

$(1 + 0) (1 - (0)) \geq 0$

Step 1.2.6.2.3

The left side $1$ 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 $x > 1$ to see if it makes the inequality true.

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Step 1.2.6.3.1

Choose a value on the interval $x > 1$ 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.

$(1 + 4) (1 - (4)) \geq 0$

Step 1.2.6.3.3

The left side $- 15$ 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.

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

Step 1.2.7

The solution consists of all of the true intervals.

$- 1 \leq x \leq 1$

Step 1.3

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

Interval Notation:

$[- 1, 1]$

Set-Builder Notation:

${x | - 1 \leq x \leq 1}$

Interval Notation:

$[- 1, 1]$

Set-Builder Notation:

${x | - 1 \leq x \leq 1}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = x \sqrt{(1 + x) (1 - x)}$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

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

Step 2.2.2

Subtract $1$ from $1$ .

$f (- 1) = - 1 \sqrt{0 (1 - (- 1))}$

Step 2.2.3

Multiply $- 1$ by $- 1$ .

$f (- 1) = - 1 \sqrt{0 (1 + 1)}$

Step 2.2.4

Add $1$ and $1$ .

$f (- 1) = - 1 \sqrt{0 \cdot 2}$

Step 2.2.5

Multiply $0$ by $2$ .

$f (- 1) = - 1 \sqrt{0}$

Step 2.2.6

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

$f (- 1) = - 1 \sqrt{0^{2}}$

Step 2.2.7

Multiply by zero.

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Step 2.2.7.1

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

$f (- 1) = - 1 \cdot 0$

Step 2.2.7.2

Multiply $- 1$ by $0$ .

$f (- 1) = 0$

Step 2.2.8

The final answer is $0$ .

$0$

Step 2.3

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

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

Step 2.4

Simplify the result.

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Step 2.4.1

Remove parentheses.

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

Step 2.4.2

Multiply $\sqrt{(1 + 1) (1 - (1))}$ by $1$ .

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

Step 2.4.3

Add $1$ and $1$ .

$f (1) = \sqrt{2 (1 - (1))}$

Step 2.4.4

Multiply $- 1$ by $1$ .

$f (1) = \sqrt{2 (1 - 1)}$

Step 2.4.5

Subtract $1$ from $1$ .

$f (1) = \sqrt{2 \cdot 0}$

Step 2.4.6

Multiply $2$ by $0$ .

$f (1) = \sqrt{0}$

Step 2.4.7

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

$f (1) = \sqrt{0^{2}}$

Step 2.4.8

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

$f (1) = 0$

Step 2.4.9

The final answer is $0$ .

$0$

Step 3

The end points are $(- 1, 0), (1, 0)$ .

$(- 1, 0), (1, 0)$

Step 4

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

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

Step 5