Find Domain and Range of Arccosine Functions

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

Solution:

• 1. Domain: To find the domain, we impose the standard condition on the argument \( (x + 1) \) according to the domain of \( \arccos(x) \) (\( -1 \le x \le 1 \)): \[ -1 \le x + 1 \le 1 \] Subtract \( 1 \) from all parts: \[ -2 \le x \le 0 \] The domain is \( [-2, 0] \). This represents a horizontal shift of the graph of \( y = \arccos(x) \) by one unit to the left.
• 2. Range: A horizontal shift does not affect the range. Hence, the range of \( y = \arccos(x + 1) \) remains identical to the range of \( \arccos(x) \): \[ 0 \le y \le \pi \quad (\text{or interval } [0, \pi]) \]

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

Solution:

• 1. Domain: Impose the condition on the argument \( (x - 2) \): \[ -1 \le x - 2 \le 1 \] Add \( 2 \) to all parts: \[ 1 \le x \le 3 \] The domain is \( [1, 3] \), representing a horizontal shift of two units to the right.
• 2. Range: Start with the standard range of \( \arccos(x - 2) \): \[ 0 \le \arccos(x - 2) \le \pi \] Multiply all terms by \( -1 \) (which reverses inequality signs): \[ 0 \ge -\arccos(x - 2) \ge -\pi \implies -\pi \le -\arccos(x - 2) \le 0 \] The range is \( [-\pi, 0] \).

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

Solution:

• 1. Domain: Impose the condition on the argument \( (-2x + 1) \): \[ -1 \le -2x + 1 \le 1 \] Subtract \( 1 \) from all parts: \[ -2 \le -2x \le 0 \] Divide by \( -2 \) and reverse inequality signs: \[ 0 \le x \le 1 \] The domain is \( [0, 1] \).
• 2. Range: Start with the base range of the arccos term: \[ 0 \le \arccos(-2x + 1) \le \pi \] Multiply all terms by \( -2 \) (reversing inequality signs): \[ 0 \ge -2\arccos(-2x + 1) \ge -2\pi \implies -2\pi \le y \le 0 \] The range is \( [-2\pi, 0] \).

Problem: Find the domain and range of \( y = 2 \arccos(2x) + \dfrac{\pi}{2} \).

Solution:

• 1. Domain: Condition on the argument \( 2x \): \[ -1 \le 2x \le 1 \implies -\dfrac{1}{2} \le x \le \dfrac{1}{2} \] The domain is \( \left[-\dfrac{1}{2}, \dfrac{1}{2}\right] \).
• 2. Range: Start with the range of \( \arccos(2x) \): \[ 0 \le \arccos(2x) \le \pi \] Multiply by \( 2 \): \[ 0 \le 2\arccos(2x) \le 2\pi \] Add \( \dfrac{\pi}{2} \) to all parts: \[ \dfrac{\pi}{2} \le 2\arccos(2x) + \dfrac{\pi}{2} \le 2\pi + \dfrac{\pi}{2} \implies \dfrac{\pi}{2} \le y \le \dfrac{5\pi}{2} \] The range is \( \left[\dfrac{\pi}{2}, \dfrac{5\pi}{2}\right] \).