Finite Math Examples

$f (x) = \sqrt{x}$ , $g (x) = \sqrt{4 - x^{2}}$

Step 1

Find the quotient of the functions.

Tap for more steps...

Step 1.1

Replace the function designators with the actual functions in $\frac{f (x)}{g (x)}$ .

$\frac{\sqrt{x}}{\sqrt{4 - x^{2}}}$

Step 1.2

Simplify.

Tap for more steps...

Step 1.2.1

Simplify the denominator.

Tap for more steps...

Step 1.2.1.1

Rewrite $4$ as $2^{2}$ .

$\frac{\sqrt{x}}{\sqrt{2^{2} - x^{2}}}$

Step 1.2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$\frac{\sqrt{x}}{\sqrt{(2 + x) (2 - x)}}$

Step 1.2.2

Multiply $\frac{\sqrt{x}}{\sqrt{(2 + x) (2 - x)}}$ by $\frac{\sqrt{(2 + x) (2 - x)}}{\sqrt{(2 + x) (2 - x)}}$ .

$\frac{\sqrt{x}}{\sqrt{(2 + x) (2 - x)}} \cdot \frac{\sqrt{(2 + x) (2 - x)}}{\sqrt{(2 + x) (2 - x)}}$

Step 1.2.3

Combine and simplify the denominator.

Tap for more steps...

Step 1.2.3.1

Multiply $\frac{\sqrt{x}}{\sqrt{(2 + x) (2 - x)}}$ by $\frac{\sqrt{(2 + x) (2 - x)}}{\sqrt{(2 + x) (2 - x)}}$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{\sqrt{(2 + x) (2 - x)} \sqrt{(2 + x) (2 - x)}}$

Step 1.2.3.2

Raise $\sqrt{(2 + x) (2 - x)}$ to the power of $1$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{\sqrt{(2 + x) (2 - x)}}^{1} \sqrt{(2 + x) (2 - x)}}$

Step 1.2.3.3

Raise $\sqrt{(2 + x) (2 - x)}$ to the power of $1$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{\sqrt{(2 + x) (2 - x)}}^{1} {\sqrt{(2 + x) (2 - x)}}^{1}}$

Step 1.2.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{\sqrt{(2 + x) (2 - x)}}^{1 + 1}}$

Step 1.2.3.5

Add $1$ and $1$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{\sqrt{(2 + x) (2 - x)}}^{2}}$

Step 1.2.3.6

Rewrite ${\sqrt{(2 + x) (2 - x)}}^{2}$ as $(2 + x) (2 - x)$ .

Tap for more steps...

Step 1.2.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(2 + x) (2 - x)}$ as ${((2 + x) (2 - x))}^{\frac{1}{2}}$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{({((2 + x) (2 - x))}^{\frac{1}{2}})}^{2}}$

Step 1.2.3.6.2

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

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{((2 + x) (2 - x))}^{\frac{1}{2} \cdot 2}}$

Step 1.2.3.6.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{((2 + x) (2 - x))}^{\frac{2}{2}}}$

Step 1.2.3.6.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.2.3.6.4.1

Cancel the common factor.

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{((2 + x) (2 - x))}^{\frac{2}{2}}}$

Step 1.2.3.6.4.2

Rewrite the expression.

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{{((2 + x) (2 - x))}^{1}}$

Step 1.2.3.6.5

Simplify.

$\frac{\sqrt{x} \sqrt{(2 + x) (2 - x)}}{(2 + x) (2 - x)}$

Step 1.2.4

Combine using the product rule for radicals.

$\frac{\sqrt{x (2 + x) (2 - x)}}{(2 + x) (2 - x)}$

Step 2

Set the radicand in $\sqrt{x (2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

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

$x = 0$

$2 + x = 0$

$2 - x = 0$

Step 3.2

Set $x$ equal to $0$ .

$x = 0$

Step 3.3

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

Tap for more steps...

Step 3.3.1

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

$2 + x = 0$

Step 3.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.4

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

Tap for more steps...

Step 3.4.1

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

$2 - x = 0$

Step 3.4.2

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

Tap for more steps...

Step 3.4.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 3.4.2.2

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

Tap for more steps...

Step 3.4.2.2.1

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

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

Step 3.4.2.2.2

Simplify the left side.

Tap for more steps...

Step 3.4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 3.4.2.2.2.2

Divide $x$ by $1$ .

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

Step 3.4.2.2.3

Simplify the right side.

Tap for more steps...

Step 3.4.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 3.5

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

$x = 0, - 2, 2$

Step 3.6

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 2$

$x > 2$

Step 3.7

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

Tap for more steps...

Step 3.7.1

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

Tap for more steps...

Step 3.7.1.1

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

$x = - 4$

Step 3.7.1.2

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

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

Step 3.7.1.3

The left side $48$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.7.2

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

Tap for more steps...

Step 3.7.2.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 3.7.2.2

Replace $x$ with $- 1$ in the original inequality.

$- 1 (2 - 1) (2 - (- 1)) \geq 0$

Step 3.7.2.3

The left side $- 3$ is less than the right side $0$ , which means that the given statement is false.

False

Step 3.7.3

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 3.7.3.1

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Step 3.7.3.2

Replace $x$ with $1$ in the original inequality.

$1 (2 + 1) (2 - (1)) \geq 0$

Step 3.7.3.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.7.4

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

Tap for more steps...

Step 3.7.4.1

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

$x = 4$

Step 3.7.4.2

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

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

Step 3.7.4.3

The left side $- 48$ is less than the right side $0$ , which means that the given statement is false.

False

Step 3.7.5

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

$x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 2$ True

$x > 2$ False

$x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 2$ True

$x > 2$ False

Step 3.8

The solution consists of all of the true intervals.

$x \leq - 2$ or $0 \leq x \leq 2$

Step 4

Set the denominator in $\frac{\sqrt{x (2 + x) (2 - x)}}{(2 + x) (2 - x)}$ equal to $0$ to find where the expression is undefined.

$(2 + x) (2 - x) = 0$

Step 5

Solve for $x$ .

Tap for more steps...

Step 5.1

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 5.2

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

Tap for more steps...

Step 5.2.1

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

$2 + x = 0$

Step 5.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5.3

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

Tap for more steps...

Step 5.3.1

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

$2 - x = 0$

Step 5.3.2

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

Tap for more steps...

Step 5.3.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 5.3.2.2

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

Tap for more steps...

Step 5.3.2.2.1

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

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

Step 5.3.2.2.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.3.2.2.2.2

Divide $x$ by $1$ .

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

Step 5.3.2.2.3

Simplify the right side.

Tap for more steps...

Step 5.3.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 5.4

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

$x = - 2, 2$

Step 6

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

Interval Notation:

$(- \infty, - 2) \cup [0, 2)$

Set-Builder Notation:

${x | x < - 2, 0 \leq x < 2}$

Step 7