Trigonometry Examples

$\frac{(x - 1) (3 - x)}{{(x - 2)}^{2}} > 0$

Step 1

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

$x - 1 = 0$

$3 - x = 0$

${(x - 2)}^{2} = 0$

Step 2

Add $1$ to both sides of the equation.

$x = 1$

Step 3

Subtract $3$ from both sides of the equation.

$- x = - 3$

Step 4

Divide each term in $- x = - 3$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Step 4.2.2

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $- 3$ by $- 1$ .

$x = 3$

Step 5

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

$x - 2 = 0$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

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

$x = 1$

$x = 3$

$x = 2$

Step 8

Consolidate the solutions.

$x = 1, 3, 2$

Step 9

Find the domain of $\frac{(x - 1) (3 - x)}{{(x - 2)}^{2}}$ .

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Step 9.1

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

${(x - 2)}^{2} = 0$

Step 9.2

Solve for $x$ .

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Step 9.2.1

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

$x - 2 = 0$

Step 9.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 9.3

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

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

Step 10

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$2 < x < 3$

$x > 3$

Step 11

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

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Step 11.1

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

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Step 11.1.1

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

$x = 0$

Step 11.1.2

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

$\frac{((0) - 1) (3 - (0))}{{((0) - 2)}^{2}} > 0$

Step 11.1.3

The left side $- 0.75$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 11.2

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

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Step 11.2.1

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

$x = 1.5$

Step 11.2.2

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

$\frac{((1.5) - 1) (3 - (1.5))}{{((1.5) - 2)}^{2}} > 0$

Step 11.2.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.3

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

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Step 11.3.1

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

$x = 2.5$

Step 11.3.2

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

$\frac{((2.5) - 1) (3 - (2.5))}{{((2.5) - 2)}^{2}} > 0$

Step 11.3.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.4

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

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Step 11.4.1

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

$x = 6$

Step 11.4.2

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

$\frac{((6) - 1) (3 - (6))}{{((6) - 2)}^{2}} > 0$

Step 11.4.3

The left side $- 0.9375$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 11.5

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

$x < 1$ False

$1 < x < 2$ True

$2 < x < 3$ True

$x > 3$ False

$x < 1$ False

$1 < x < 2$ True

$2 < x < 3$ True

$x > 3$ False

Step 12

The solution consists of all of the true intervals.

$1 < x < 2$ or $2 < x < 3$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$1 < x < 2 or 2 < x < 3$

Interval Notation:

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

Step 14