Find the Domain and Range y=2/3arccos(3x+1/2)

$y = \frac{2}{3} \arccos (3 x + \frac{1}{2})$

Step 1

Set the argument in $\arccos (3 x + \frac{1}{2})$ greater than or equal to $- 1$ to find where the expression is defined.

$3 x + \frac{1}{2} \geq - 1$

Step 2

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $\frac{1}{2}$ from both sides of the inequality.

$3 x \geq - 1 - \frac{1}{2}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 x \geq - 1 \cdot \frac{2}{2} - \frac{1}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

$3 x \geq \frac{- 1 \cdot 2}{2} - \frac{1}{2}$

Combine the numerators over the common denominator.

$3 x \geq \frac{- 1 \cdot 2 - 1}{2}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$3 x \geq \frac{- 2 - 1}{2}$

Subtract $1$ from $- 2$ .

$3 x \geq \frac{- 3}{2}$

Move the negative in front of the fraction.

$3 x \geq - \frac{3}{2}$

Divide each term in $3 x \geq - \frac{3}{2}$ by $3$ and simplify.

Divide each term in $3 x \geq - \frac{3}{2}$ by $3$ .

$\frac{3 x}{3} \geq \frac{- \frac{3}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{- \frac{3}{2}}{3}$

Divide $x$ by $1$ .

$x \geq \frac{- \frac{3}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x \geq - \frac{3}{2} \cdot \frac{1}{3}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$x \geq \frac{- 3}{2} \cdot \frac{1}{3}$

Factor $3$ out of $- 3$ .

$x \geq \frac{3 (- 1)}{2} \cdot \frac{1}{3}$

Cancel the common factor.

$x \geq \frac{3 \cdot - 1}{2} \cdot \frac{1}{3}$

Rewrite the expression.

$x \geq \frac{- 1}{2}$

Move the negative in front of the fraction.

$x \geq - \frac{1}{2}$

Step 3

Set the argument in $\arccos (3 x + \frac{1}{2})$ less than or equal to $1$ to find where the expression is defined.

$3 x + \frac{1}{2} \leq 1$

Step 4

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $\frac{1}{2}$ from both sides of the inequality.

$3 x \leq 1 - \frac{1}{2}$

Write $1$ as a fraction with a common denominator.

$3 x \leq \frac{2}{2} - \frac{1}{2}$

Combine the numerators over the common denominator.

$3 x \leq \frac{2 - 1}{2}$

Subtract $1$ from $2$ .

$3 x \leq \frac{1}{2}$

Divide each term in $3 x \leq \frac{1}{2}$ by $3$ and simplify.

Divide each term in $3 x \leq \frac{1}{2}$ by $3$ .

$\frac{3 x}{3} \leq \frac{\frac{1}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{\frac{1}{2}}{3}$

Divide $x$ by $1$ .

$x \leq \frac{\frac{1}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x \leq \frac{1}{2} \cdot \frac{1}{3}$

Multiply $\frac{1}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{2}$ by $\frac{1}{3}$ .

$x \leq \frac{1}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x \leq \frac{1}{6}$

Step 5

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

Interval Notation:

$[- \frac{1}{2}, \frac{1}{6}]$

Set-Builder Notation:

${x | - \frac{1}{2} \leq x \leq \frac{1}{6}}$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \frac{2 π}{3}]$

Set-Builder Notation:

${y | 0 \leq y \leq \frac{2 π}{3}}$

Step 7

Determine the domain and range.

Domain: $[- \frac{1}{2}, \frac{1}{6}], {x | - \frac{1}{2} \leq x \leq \frac{1}{6}}$

Range: $[0, \frac{2 π}{3}], {y | 0 \leq y \leq \frac{2 π}{3}}$

Step 8

Find the Domain and Range y=2/3arccos(3x+1/2)

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