Find the Domain f(x)=1/(x^2-2x-3)

$f (x) = \frac{1}{x^{2} - 2 x - 3}$

Step 1

Set the denominator in $\frac{1}{x^{2} - 2 x - 3}$ equal to $0$ to find where the expression is undefined.

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

Step 2

Solve for $x$ .

Factor $x^{2} - 2 x - 3$ using the AC method.

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 $- 2$ .

$- 3, 1$

Write the factored form using these integers.

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

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

$x - 3 = 0$

$x + 1 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Find the Domain f(x)=1/(x^2-2x-3)

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