Precalculus Examples

$g (t) = \frac{\ln (π - | t |)}{\sin (t) - \frac{1}{2}}$

Step 1

Set the argument in $\ln (π - | t |)$ greater than $0$ to find where the expression is defined.

$π - | t | > 0$

Step 2

Solve for $t$ .

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Step 2.1

Write $π - | t | > 0$ as a piecewise.

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Step 2.1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$t \geq 0$

Step 2.1.2

In the piece where $t$ is non-negative, remove the absolute value.

$π - t > 0$

Step 2.1.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$t < 0$

Step 2.1.4

In the piece where $t$ is negative, remove the absolute value and multiply by $- 1$ .

$π - - t > 0$

Step 2.1.5

Write as a piecewise.

${\begin{cases} π - t > 0 & t \geq 0 \\ π - - t > 0 & t < 0 \end{cases}$

Step 2.1.6

Multiply $- - t$ .

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Step 2.1.6.1

Multiply $- 1$ by $- 1$ .

${\begin{cases} π - t > 0 & t \geq 0 \\ π + 1 t > 0 & t < 0 \end{cases}$

Step 2.1.6.2

Multiply $t$ by $1$ .

${\begin{cases} π - t > 0 & t \geq 0 \\ π + t > 0 & t < 0 \end{cases}$

Step 2.2

Solve $π - t > 0$ when $t \geq 0$ .

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Step 2.2.1

Solve $π - t > 0$ for $t$ .

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Step 2.2.1.1

Subtract $π$ from both sides of the inequality.

$- t > - π$

Step 2.2.1.2

Divide each term in $- t > - π$ by $- 1$ and simplify.

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Step 2.2.1.2.1

Divide each term in $- t > - π$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- t}{- 1} < \frac{- π}{- 1}$

Step 2.2.1.2.2

Simplify the left side.

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Step 2.2.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{t}{1} < \frac{- π}{- 1}$

Step 2.2.1.2.2.2

Divide $t$ by $1$ .

$t < \frac{- π}{- 1}$

Step 2.2.1.2.3

Simplify the right side.

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Step 2.2.1.2.3.1

Dividing two negative values results in a positive value.

$t < \frac{π}{1}$

Step 2.2.1.2.3.2

Divide $π$ by $1$ .

$t < π$

Step 2.2.2

Find the intersection of $t < π$ and $t \geq 0$ .

$0 \leq t < π$

Step 2.3

Solve $π + t > 0$ when $t < 0$ .

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Step 2.3.1

Subtract $π$ from both sides of the inequality.

$t > - π$

Step 2.3.2

Find the intersection of $t > - π$ and $t < 0$ .

$- π < t < 0$

Step 2.4

Find the union of the solutions.

$- π < t < π$

Step 3

Set the denominator in $\frac{\ln (π - | t |)}{\sin (t) - \frac{1}{2}}$ equal to $0$ to find where the expression is undefined.

$\sin (t) - \frac{1}{2} = 0$

Step 4

Solve for $t$ .

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Step 4.1

Add $\frac{1}{2}$ to both sides of the equation.

$\sin (t) = \frac{1}{2}$

Step 4.2

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

$t = \arcsin (\frac{1}{2})$

Step 4.3

Simplify the right side.

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Step 4.3.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$t = \frac{π}{6}$

Step 4.4

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.

$t = π - \frac{π}{6}$

Step 4.5

Simplify $π - \frac{π}{6}$ .

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Step 4.5.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$t = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 4.5.2

Combine fractions.

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Step 4.5.2.1

Combine $π$ and $\frac{6}{6}$ .

$t = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 4.5.2.2

Combine the numerators over the common denominator.

$t = \frac{π \cdot 6 - π}{6}$

Step 4.5.3

Simplify the numerator.

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Step 4.5.3.1

Move $6$ to the left of $π$ .

$t = \frac{6 \cdot π - π}{6}$

Step 4.5.3.2

Subtract $π$ from $6 π$ .

$t = \frac{5 π}{6}$

Step 4.6

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

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Step 4.6.1

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

$\frac{2 π}{| b |}$

Step 4.6.2

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

$\frac{2 π}{| 1 |}$

Step 4.6.3

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

$\frac{2 π}{1}$

Step 4.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.7

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

$t = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 5

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

Set-Builder Notation:

${t | - π < t < π, t \neq \frac{π}{6} + 2 π n, t \neq \frac{5 π}{6} + 2 π n}$

Step 6