Precalculus Examples

$f (x) = \frac{x^{2} - 25}{x^{2} + 5 x}$

Step 1

Set the denominator in $\frac{x^{2} - 25}{x^{2} + 5 x}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 5 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x$ out of $x^{2} + 5 x$ .

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Step 2.1.1

Factor $x$ out of $x^{2}$ .

$x \cdot x + 5 x = 0$

Step 2.1.2

Factor $x$ out of $5 x$ .

$x \cdot x + x \cdot 5 = 0$

Step 2.1.3

Factor $x$ out of $x \cdot x + x \cdot 5$ .

$x (x + 5) = 0$

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 $0$ .

$x = 0$

$x + 5 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.5

The final solution is all the values that make $x (x + 5) = 0$ true.

$x = 0, - 5$

Step 3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 5}$

Step 4

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

Interval Notation:

$(- \infty, 1) \cup (1, 2) \cup (2, \infty)$

Set-Builder Notation:

${y | y \neq 1, 2}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 5) \cup (- 5, 0) \cup (0, \infty), {x | x \neq 0, - 5}$

Range: $(- \infty, 1) \cup (1, 2) \cup (2, \infty), {y | y \neq 1, 2}$

Step 6