Calculus Examples

$f (x) = \frac{x + 2}{x^{2} - 7 x - 18}$

Step 1

Set the denominator in $\frac{x + 2}{x^{2} - 7 x - 18}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 7 x - 18 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - 7 x - 18$ using the AC method.

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Step 2.1.1

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 $- 18$ and whose sum is $- 7$ .

$- 9, 2$

Step 2.1.2

Write the factored form using these integers.

$(x - 9) (x + 2) = 0$

Step 2.2

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

$x - 9 = 0$

$x + 2 = 0$

Step 2.3

Set $x - 9$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Step 2.3.2

Add $9$ to both sides of the equation.

$x = 9$

Step 2.4

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

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Step 2.4.1

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

$x + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5

The final solution is all the values that make $(x - 9) (x + 2) = 0$ true.

$x = 9, - 2$

Step 3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 9) \cup (9, \infty)$

Set-Builder Notation:

${x | x \neq - 2, 9}$

Step 4

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

Interval Notation:

$(- \infty, - \frac{1}{11}) \cup (- \frac{1}{11}, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0, - \frac{1}{11}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 2) \cup (- 2, 9) \cup (9, \infty), {x | x \neq - 2, 9}$

Range: $(- \infty, - \frac{1}{11}) \cup (- \frac{1}{11}, 0) \cup (0, \infty), {y | y \neq 0, - \frac{1}{11}}$

Step 6