Find the Domain and Range y=arcsec(x+1)

$y = arcsec (x + 1)$

Set the argument in $arcsec (x + 1)$ less than or equal to $- 1$ to find where the expression is defined.

$x + 1 \leq - 1$

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

Subtract $1$ from both sides of the inequality.

$x \leq - 1 - 1$

Subtract $1$ from $- 1$ .

$x \leq - 2$

Set the argument in $arcsec (x + 1)$ greater than or equal to $1$ to find where the expression is defined.

$x + 1 \geq 1$

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

Subtract $1$ from both sides of the inequality.

$x \geq 1 - 1$

Subtract $1$ from $1$ .

$x \geq 0$

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

Interval Notation:

$(- \infty, - 2] \cup [0, \infty)$

Set-Builder Notation:

${x | x \leq - 2, x \geq 0}$

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: $(- \infty, - 2] \cup [0, \infty), {x | x \leq - 2, x \geq 0}$

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