Precalculus Examples

$f (x) = \sqrt{\frac{x - 1}{x^{2} - 5 x + 6}}$

Step 1

Set the radicand in $\sqrt{\frac{x - 1}{x^{2} - 5 x + 6}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{x - 1}{x^{2} - 5 x + 6} \geq 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 1 = 0$

$x^{2} - 5 x + 6 = 0$

Step 2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.3

Factor $x^{2} - 5 x + 6$ using the AC method.

Tap for more steps...

Step 2.3.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 2.3.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 2.4

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

$x - 3 = 0$

$x - 2 = 0$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$x - 3 = 0$

Step 2.5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.6

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

Tap for more steps...

Step 2.6.1

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

$x - 2 = 0$

Step 2.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 2.8

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1$

$x = 3, 2$

Step 2.9

Consolidate the solutions.

$x = 1, 3, 2$

Step 2.10

Find the domain of $\frac{x - 1}{x^{2} - 5 x + 6}$ .

Tap for more steps...

Step 2.10.1

Set the denominator in $\frac{x - 1}{x^{2} - 5 x + 6}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x + 6 = 0$

Step 2.10.2

Solve for $x$ .

Tap for more steps...

Step 2.10.2.1

Factor $x^{2} - 5 x + 6$ using the AC method.

Tap for more steps...

Step 2.10.2.1.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 2.10.2.1.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 2.10.2.2

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

$x - 3 = 0$

$x - 2 = 0$

Step 2.10.2.3

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

Tap for more steps...

Step 2.10.2.3.1

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

$x - 3 = 0$

Step 2.10.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.10.2.4

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

Tap for more steps...

Step 2.10.2.4.1

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

$x - 2 = 0$

Step 2.10.2.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.10.2.5

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 2.10.3

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

$(- \infty, 2) \cup (2, 3) \cup (3, \infty)$

Step 2.11

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$2 < x < 3$

$x > 3$

Step 2.12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 2.12.1

Test a value on the interval $x < 1$ to see if it makes the inequality true.

Tap for more steps...

Step 2.12.1.1

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.12.1.2

Replace $x$ with $0$ in the original inequality.

$\frac{(0) - 1}{{(0)}^{2} - 5 \cdot 0 + 6} \geq 0$

Step 2.12.1.3

The left side $- 0.1\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.12.2

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 2.12.2.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 2.12.2.2

Replace $x$ with $1.5$ in the original inequality.

$\frac{(1.5) - 1}{{(1.5)}^{2} - 5 \cdot 1.5 + 6} \geq 0$

Step 2.12.2.3

The left side $0.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.3

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Tap for more steps...

Step 2.12.3.1

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Step 2.12.3.2

Replace $x$ with $2.5$ in the original inequality.

$\frac{(2.5) - 1}{{(2.5)}^{2} - 5 \cdot 2.5 + 6} \geq 0$

Step 2.12.3.3

The left side $- 6$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.12.4

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Tap for more steps...

Step 2.12.4.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 2.12.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{(6) - 1}{{(6)}^{2} - 5 \cdot 6 + 6} \geq 0$

Step 2.12.4.3

The left side $0.41\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < 1$ False

$1 < x < 2$ True

$2 < x < 3$ False

$x > 3$ True

$x < 1$ False

$1 < x < 2$ True

$2 < x < 3$ False

$x > 3$ True

Step 2.13

The solution consists of all of the true intervals.

$1 \leq x < 2$ or $x > 3$

Step 3

Set the denominator in $\frac{x - 1}{x^{2} - 5 x + 6}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x + 6 = 0$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

Factor $x^{2} - 5 x + 6$ using the AC method.

Tap for more steps...

Step 4.1.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 4.1.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 4.2

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

$x - 3 = 0$

$x - 2 = 0$

Step 4.3

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

Tap for more steps...

Step 4.3.1

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

$x - 3 = 0$

Step 4.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.4

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

Tap for more steps...

Step 4.4.1

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

$x - 2 = 0$

Step 4.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.5

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 5

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

Interval Notation:

$[1, 2) \cup (3, \infty)$

Set-Builder Notation:

${x | 1 \leq x < 2, x > 3}$

Step 6