Find the Domain and Range y=arcsec(7x-1/2)

$y = arcsec (7 x - \frac{1}{2})$

Set the argument in $arcsec (7 x - \frac{1}{2})$ less than or equal to $- 1$ to find where the expression is defined.

$7 x - \frac{1}{2} \leq - 1$

Solve for $x$ .

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

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

$7 x \leq - 1 + \frac{1}{2}$

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

$7 x \leq - 1 \cdot \frac{2}{2} + \frac{1}{2}$

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

$7 x \leq \frac{- 1 \cdot 2}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$7 x \leq \frac{- 1 \cdot 2 + 1}{2}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$7 x \leq \frac{- 2 + 1}{2}$

Add $- 2$ and $1$ .

$7 x \leq \frac{- 1}{2}$

Move the negative in front of the fraction.

$7 x \leq - \frac{1}{2}$

Divide each term in $7 x \leq - \frac{1}{2}$ by $7$ and simplify.

Divide each term in $7 x \leq - \frac{1}{2}$ by $7$ .

$\frac{7 x}{7} \leq \frac{- \frac{1}{2}}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} \leq \frac{- \frac{1}{2}}{7}$

Divide $x$ by $1$ .

$x \leq \frac{- \frac{1}{2}}{7}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x \leq - \frac{1}{2} \cdot \frac{1}{7}$

Multiply $- \frac{1}{2} \cdot \frac{1}{7}$ .

Multiply $\frac{1}{7}$ by $\frac{1}{2}$ .

$x \leq - \frac{1}{7 \cdot 2}$

Multiply $7$ by $2$ .

$x \leq - \frac{1}{14}$

Set the argument in $arcsec (7 x - \frac{1}{2})$ greater than or equal to $1$ to find where the expression is defined.

$7 x - \frac{1}{2} \geq 1$

Solve for $x$ .

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

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

$7 x \geq 1 + \frac{1}{2}$

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

$7 x \geq \frac{2}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$7 x \geq \frac{2 + 1}{2}$

Add $2$ and $1$ .

$7 x \geq \frac{3}{2}$

Divide each term in $7 x \geq \frac{3}{2}$ by $7$ and simplify.

Divide each term in $7 x \geq \frac{3}{2}$ by $7$ .

$\frac{7 x}{7} \geq \frac{\frac{3}{2}}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} \geq \frac{\frac{3}{2}}{7}$

Divide $x$ by $1$ .

$x \geq \frac{\frac{3}{2}}{7}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x \geq \frac{3}{2} \cdot \frac{1}{7}$

Multiply $\frac{3}{2} \cdot \frac{1}{7}$ .

Multiply $\frac{3}{2}$ by $\frac{1}{7}$ .

$x \geq \frac{3}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x \geq \frac{3}{14}$

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

Interval Notation:

$(- \infty, - \frac{1}{14}] \cup [\frac{3}{14}, \infty)$

Set-Builder Notation:

${x | x \leq - \frac{1}{14}, x \geq \frac{3}{14}}$

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

Interval Notation:

$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$

Set-Builder Notation:

${y | 0 \leq y \leq π, y \neq \frac{π}{2}}$

Determine the domain and range.

Domain: $(- \infty, - \frac{1}{14}] \cup [\frac{3}{14}, \infty), {x | x \leq - \frac{1}{14}, x \geq \frac{3}{14}}$

Range: $[0, \frac{π}{2}) \cup (\frac{π}{2}, π], {y | 0 \leq y \leq π, y \neq \frac{π}{2}}$

Find the Domain and Range y=arcsec(7x-1/2)

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