Linear Algebra Examples

$\sqrt{\frac{\sin (x)}{x}}$

Step 1

Set the radicand in $\sqrt{\frac{\sin (x)}{x}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{\sin (x)}{x} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x = 0$

$\sin (x) = 0$

Step 2.2

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (0)$

Step 2.3

Simplify the right side.

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Step 2.3.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 2.4

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - 0$

Step 2.5

Subtract $0$ from $π$ .

$x = π$

Step 2.6

Find the period of $\sin (x)$ .

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Step 2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.7

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 2.8

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

$x = 0$

$x = 2 π n, π + 2 π n$

Step 2.9

Consolidate the solutions.

$x = 0, 2 π n, π + 2 π n$

Step 2.10

Find the domain of $\frac{\sin (x)}{x}$ .

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Step 2.10.1

Set the denominator in $\frac{\sin (x)}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 2.10.2

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

$(- \infty, 0) \cup (0, \infty)$

Step 2.11

Use each root to create test intervals.

$0 < x < π$

$π < x < 2 π$

$2 π < x < 3 π$

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.

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Step 2.12.1

Test a value on the interval $0 < x < π$ to see if it makes the inequality true.

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Step 2.12.1.1

Choose a value on the interval $0 < x < π$ and see if this value makes the original inequality true.

$x = 2$

Step 2.12.1.2

Replace $x$ with $2$ in the original inequality.

$\frac{\sin (2)}{2} \geq 0$

Step 2.12.1.3

The left side $0.45464871$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.2

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

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Step 2.12.2.1

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

$x = 5$

Step 2.12.2.2

Replace $x$ with $5$ in the original inequality.

$\frac{\sin (5)}{5} \geq 0$

Step 2.12.2.3

The left side $- 0.19178485$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.12.3

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

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Step 2.12.3.1

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

$x = 8$

Step 2.12.3.2

Replace $x$ with $8$ in the original inequality.

$\frac{\sin (8)}{8} \geq 0$

Step 2.12.3.3

The left side $0.12366978$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.4

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

$0 < x < π$ True

$π < x < 2 π$ False

$2 π < x < 3 π$ True

$0 < x < π$ True

$π < x < 2 π$ False

$2 π < x < 3 π$ True

Step 2.13

The solution consists of all of the true intervals.

$0 + 2 π n \leq x \leq π + 2 π n$ or $2 π + 2 π n \leq x \leq 3 π + 2 π n$ , for any integer $n$

Step 2.14

Combine the intervals.

$2 π n \leq x \leq π + 2 π n or 2 π + 2 π n \leq x \leq 3 π + 2 π n$ , for any integer $n$

Step 3

Set the denominator in $\frac{\sin (x)}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 4

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

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 5