Finite Math Examples

$f (x) = \frac{\sqrt{x + 2}}{x^{2} - 2 x - 8}$

Step 1

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

$x + 2 \geq 0$

Step 2

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Step 3

Set the denominator in $\frac{\sqrt{x + 2}}{x^{2} - 2 x - 8}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 2 x - 8 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor $x^{2} - 2 x - 8$ using the AC method.

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Step 4.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Step 4.1.2

Write the factored form using these integers.

$(x - 4) (x + 2) = 0$

Step 4.2

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

$x - 4 = 0$

$x + 2 = 0$

Step 4.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 4.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 4.4

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

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Step 4.4.1

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

$x + 2 = 0$

Step 4.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.5

The final solution is all the values that make $(x - 4) (x + 2) = 0$ true.

$x = 4, - 2$

Step 5

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

Interval Notation:

$(- 2, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x > - 2, x \neq 4}$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - \frac{\sqrt{2}}{8}] \cup (0, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{\sqrt{2}}{8}, y > 0}$

Step 7

Determine the domain and range.

Domain: $(- 2, 4) \cup (4, \infty), {x | x > - 2, x \neq 4}$

Range: $(- \infty, - \frac{\sqrt{2}}{8}] \cup (0, \infty), {y | y \leq - \frac{\sqrt{2}}{8}, y > 0}$

Step 8