Precalculus Examples

$f (x) = \frac{2}{x^{2} - 2 x - 3}$

Step 1

Set the denominator in $\frac{2}{x^{2} - 2 x - 3}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 2 x - 3 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 2.1.2

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

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 $0$ .

$x - 3 = 0$

$x + 1 = 0$

Step 2.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.4

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq - 1, 3}$

Step 4

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

Interval Notation:

$(- \infty, - \frac{1}{2}] \cup (0, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{1}{2}, y > 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 1) \cup (- 1, 3) \cup (3, \infty), {x | x \neq - 1, 3}$

Range: $(- \infty, - \frac{1}{2}] \cup (0, \infty), {y | y \leq - \frac{1}{2}, y > 0}$

Step 6