Algebra Examples

f (x) = \frac{1}{x - 5}

$f (x) = \frac{1}{x - 5}$

Step 1

Set the denominator in

\frac{1}{x - 5}

$\frac{1}{x - 5}$ equal to

0

$0$ to find where the expression is undefined.

x - 5 = 0

$x - 5 = 0$

Step 2

Add

5

$5$ to both sides of the equation.

x = 5

$x = 5$

Step 3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

(- \infty, 5) \cup (5, \infty)

$(- \infty, 5) \cup (5, \infty)$

Set-Builder Notation:

{x | x \neq 5}

${x | x \neq 5}$

Step 4

The range is the set of all valid

y

$y$ values. Use the graph to find the range.

Interval Notation:

(- \infty, 0) \cup (0, \infty)

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

Set-Builder Notation:

{y | y \neq 0}

${y | y \neq 0}$

Step 5

Determine the domain and range.

Domain:

(- \infty, 5) \cup (5, \infty), {x | x \neq 5}

$(- \infty, 5) \cup (5, \infty), {x | x \neq 5}$

Range:

(- \infty, 0) \cup (0, \infty), {y | y \neq 0}

$(- \infty, 0) \cup (0, \infty), {y | y \neq 0}$

Step 6

f (x) = \frac{1}{x - 5}

$f (x) = \frac{1}{x - 5}$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$