Precalculus Examples

$k (x) = \frac{1}{(x + 5) (x - 2) (x - 7)}$

Step 1

Set the denominator in $\frac{1}{(x + 5) (x - 2) (x - 7)}$ equal to $0$ to find where the expression is undefined.

$(x + 5) (x - 2) (x - 7) = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x + 5 = 0$

$x - 2 = 0$

$x - 7 = 0$

Step 2.2

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

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Step 2.2.1

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

$x + 5 = 0$

Step 2.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3

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

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Step 2.3.1

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

$x - 2 = 0$

Step 2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.4

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.5

The final solution is all the values that make $(x + 5) (x - 2) (x - 7) = 0$ true.

$x = - 5, 2, 7$

Step 3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 2) \cup (2, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 5, 7}$

Step 4