Precalculus Examples

$f (x) = \sqrt{3 x - 5}$ , $g (x) = x^{2} - 4 x - 12$

Step 1

Find the quotient of the functions.

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Step 1.1

Replace the function designators with the actual functions in $\frac{f (x)}{g (x)}$ .

$\frac{\sqrt{3 x - 5}}{x^{2} - 4 x - 12}$

Step 1.2

Factor $x^{2} - 4 x - 12$ using the AC method.

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Step 1.2.1

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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 1.2.2

Write the factored form using these integers.

$\frac{\sqrt{3 x - 5}}{(x - 6) (x + 2)}$

Step 2

Set the radicand in $\sqrt{3 x - 5}$ greater than or equal to $0$ to find where the expression is defined.

$3 x - 5 \geq 0$

Step 3

Solve for $x$ .

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Step 3.1

Add $5$ to both sides of the inequality.

$3 x \geq 5$

Step 3.2

Divide each term in $3 x \geq 5$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 x \geq 5$ by $3$ .

$\frac{3 x}{3} \geq \frac{5}{3}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{5}{3}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{5}{3}$

Step 4

Set the denominator in $\frac{\sqrt{3 x - 5}}{(x - 6) (x + 2)}$ equal to $0$ to find where the expression is undefined.

$(x - 6) (x + 2) = 0$

Step 5

Solve for $x$ .

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Step 5.1

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

$x - 6 = 0$

$x + 2 = 0$

Step 5.2

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 5.2.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 5.2.2

Add $6$ to both sides of the equation.

$x = 6$

Step 5.3

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

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Step 5.3.1

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

$x + 2 = 0$

Step 5.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5.4

The final solution is all the values that make $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Step 6

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

Interval Notation:

$[\frac{5}{3}, 6) \cup (6, \infty)$

Set-Builder Notation:

${x | x \geq \frac{5}{3}, x \neq 6}$

Step 7