Finite Math Examples

$f (x) = \frac{\sqrt{x + 6}}{x^{2} - x - 42}$

Step 1

Set the radicand in $\sqrt{x + 6}$ greater than or equal to $0$ to find where the expression is defined.

$x + 6 \geq 0$

Step 2

Subtract $6$ from both sides of the inequality.

$x \geq - 6$

Step 3

Set the denominator in $\frac{\sqrt{x + 6}}{x^{2} - x - 42}$ equal to $0$ to find where the expression is undefined.

$x^{2} - x - 42 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor $x^{2} - x - 42$ 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 $- 42$ and whose sum is $- 1$ .

$- 7, 6$

Step 4.1.2

Write the factored form using these integers.

$(x - 7) (x + 6) = 0$

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 $0$ .

$x - 7 = 0$

$x + 6 = 0$

Step 4.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 4.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 4.4

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 4.4.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 4.5

The final solution is all the values that make $(x - 7) (x + 6) = 0$ true.

$x = 7, - 6$

Step 5

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

Interval Notation:

$(- 6, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x > - 6, x \neq 7}$

Step 6