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

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

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

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

Solve for $x$ .

Multiply both sides by $2$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$x + 1 \geq - 1 \cdot 2$

Simplify the right side.

Multiply $- 1$ by $2$ .

$x + 1 \geq - 2$

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

Subtract $1$ from both sides of the inequality.

$x \geq - 2 - 1$

Subtract $1$ from $- 2$ .

$x \geq - 3$

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

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

Solve for $x$ .

Multiply both sides by $2$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$x + 1 \leq 1 \cdot 2$

Simplify the right side.

Multiply $2$ by $1$ .

$x + 1 \leq 2$

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

Subtract $1$ from both sides of the inequality.

$x \leq 2 - 1$

Subtract $1$ from $2$ .

$x \leq 1$

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

Interval Notation:

$[- 3, 1]$

Set-Builder Notation:

${x | - 3 \leq x \leq 1}$

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

Interval Notation:

$[- \frac{π}{2}, \frac{π}{2}]$

Set-Builder Notation:

${y | - \frac{π}{2} \leq y \leq \frac{π}{2}}$

Determine the domain and range.

Domain: $[- 3, 1], {x | - 3 \leq x \leq 1}$

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

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

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