Finding Domain and Range of Arcsine Functions

Problem: Find the domain and range of \( y = \arcsin(x - 1) \).

Solution:

• 1. 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 by adding \(1\) to all parts: \[ 0 \le x \le 2 \] The domain is \([0, 2]\). This shows that the graph of \( y = \arcsin(x - 1) \) is the graph of \( y = \arcsin(x) \) shifted one unit to the right.
• 2. Range: Horizontal shifts do not affect the range. Therefore, the range remains: \[ -\frac{\pi}{2} \le y \le \frac{\pi}{2} \quad (\text{or interval } \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]) \]

Problem: Find the domain and range of \( y = -\arcsin(x + 2) \).

Solution:

• 1. Domain: Apply the domain condition to \((x + 2)\): \[ -1 \le x + 2 \le 1 \] Subtract \(2\) from all parts: \[ -3 \le x \le -1 \] The domain is \([-3, -1]\), representing a horizontal shift two units to the left.
• 2. Range: Start with the standard range of \(\arcsin(x + 2)\): \[ -\frac{\pi}{2} \le \arcsin(x + 2) \le \frac{\pi}{2} \] Multiply by \(-1\) (which reverses 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 \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Problem: Find the domain and range of \( y = -2\arcsin(3x - 1) \).

Solution:

• 1. Domain: Apply the domain condition to \((3x - 1)\): \[ -1 \le 3x - 1 \le 1 \] Add \(1\) to all parts: \[ 0 \le 3x \le 2 \] Divide by \(3\): \[ 0 \le x \le \frac{2}{3} \] The domain is \(\left[0, \frac{2}{3}\right]\).
• 2. Range: Start with the base range of the arcsine function: \[ -\frac{\pi}{2} \le \arcsin(3x - 1) \le \frac{\pi}{2} \] Multiply by \(-2\) (reversing inequality signs): \[ \pi \ge -2\arcsin(3x - 1) \ge -\pi \implies -\pi \le y \le \pi \] Thus, the range is \([-\pi, \pi]\).

Problem: Find the domain and range of \( y = 4\arcsin(-2(x - 1)) - \dfrac{\pi}{2} \).

Solution:

• 1. Domain: Apply the domain condition to \((-2(x - 1))\): \[ -1 \le -2(x - 1) \le 1 \] Divide by \(-2\) (reversing inequality signs): \[ \frac{1}{2} \ge x - 1 \ge -\frac{1}{2} \implies -\frac{1}{2} \le x - 1 \le \frac{1}{2} \] Add \(1\) to all parts: \[ \frac{1}{2} \le x \le \frac{3}{2} \] The domain is \(\left[\frac{1}{2}, \frac{3}{2}\right]\).
• 2. Range: Start with the base range: \[ -\frac{\pi}{2} \le \arcsin(-2(x - 1)) \le \frac{\pi}{2} \] Multiply by \(4\): \[ -2\pi \le 4\arcsin(-2(x - 1)) \le 2\pi \] Subtract \(\dfrac{\pi}{2}\) from all parts: \[ -2\pi - \frac{\pi}{2} \le 4\arcsin(-2(x - 1)) - \frac{\pi}{2} \le 2\pi - \frac{\pi}{2} \implies -\frac{5\pi}{2} \le y \le \frac{3\pi}{2} \] Thus, the range is \(\left[-\frac{5\pi}{2}, \frac{3\pi}{2}\right]\).