Learn how to determine domain and range of arcsine functions with detailed step-by-step solutions.
Theorem
The arcsine function is defined as:
\[ y = \arcsin(x) \quad \text{is equivalent to} \quad \sin(y) = x \]with the constraints:
\[ -1 \le x \le 1 \quad \text{and} \quad -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]Question 1
Find the domain and range of \( y = \arcsin(x - 1) \).
Solution
Domain: We impose the condition on the argument \((x - 1)\) based on the domain of \(\arcsin(x)\):
\[ -1 \le (x - 1) \le 1 \]Solving the inequality:
\[ 0 \le x \le 2 \]This shows the graph of \( y = \arcsin(x - 1) \) is the graph of \( y = \arcsin(x) \) shifted one unit to the right.
Range: Horizontal shifts do not affect the range. Therefore:
\[ -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]Question 2
Find the domain and range of \( y = -\arcsin(x + 2) \).
Solution
Domain: Apply the domain condition to \((x + 2)\):
\[ -1 \le (x + 2) \le 1 \]Solving:
\[ -3 \le x \le -1 \]This represents a horizontal shift two units to the left.
Range: Start with the range of \(\arcsin(x + 2)\):
\[ -\frac{\pi}{2} \le \arcsin(x + 2) \le \frac{\pi}{2} \]Multiply by \(-1\) (reverse inequality signs):
\[ \frac{\pi}{2} \ge -\arcsin(x + 2) \ge -\frac{\pi}{2} \]Rewriting in standard order:
\[ -\frac{\pi}{2} \le -\arcsin(x + 2) \le \frac{\pi}{2} \]Thus the range is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Question 3
Find the domain and range of \( y = -2\arcsin(3x - 1) \).
Solution
Domain: Apply domain condition to \((3x - 1)\):
\[ -1 \le (3x - 1) \le 1 \]Solving:
\[ 0 \le x \le \frac{2}{3} \]Range: Start with:
\[ -\frac{\pi}{2} \le \arcsin(3x - 1) \le \frac{\pi}{2} \]Multiply by \(-2\) (reverse inequality signs):
\[ \pi \ge -2\arcsin(3x - 1) \ge -\pi \]Thus the range is \([-\pi, \pi]\).
Question 4
Find the domain and range of \( y = 4\arcsin(-2(x - 1)) - \frac{\pi}{2} \).
Solution
Domain: Apply domain condition to \((-2(x - 1))\):
\[ -1 \le -2(x - 1) \le 1 \]Solving:
\[ \frac{1}{2} \le x \le \frac{3}{2} \]Range: Start with:
\[ -\frac{\pi}{2} \le \arcsin(-2(x - 1)) \le \frac{\pi}{2} \]Multiply by 4:
\[ -2\pi \le 4\arcsin(-2(x - 1)) \le 2\pi \]Subtract \(\frac{\pi}{2}\):
\[ -\frac{5\pi}{2} \le 4\arcsin(-2(x - 1)) - \frac{\pi}{2} \le \frac{3\pi}{2} \]Thus the range is \(\left[-\frac{5\pi}{2}, \frac{3\pi}{2}\right]\).