Questions on how to find domain and range of arccosine functions with detailed solutions.
\[ y = \arccos x \quad \text{is equivalent to} \quad \cos y = x \]
with \( -1 \le x \le 1 \quad \text{and} \quad 0 \le y \le \pi \)
Find the domain and range of \( y = \arccos(x + 1) \)
1. Domain: To find the domain, we need to impose a condition on the argument \( (x + 1) \) according to the domain of \( \arccos(x) \) which is \( -1 \le x \le 1 \). Hence
\[ -1 \le (x + 1) \le 1 \]
Solve to obtain domain as: \( -2 \le x \le 0 \)
This means the graph of \( y = \arccos(x + 1) \) is that of \( y = \arccos(x) \) shifted one unit to the left.
2. Range: A horizontal shift does not affect the range. Hence the range of \( y = \arccos(x + 1) \) is the same as the range of \( \arccos(x) \), which is \( 0 \le y \le \pi \).
Find the domain and range of \( y = -\arccos(x - 2) \)
1. Domain: We impose the condition on \( (x - 2) \) according to the domain of \( \arccos(x) \):
\[ -1 \le (x - 2) \le 1 \]
Solve to obtain domain: \( 1 \le x \le 3 \)
This represents a horizontal shift two units to the right.
2. Range: The range of \( \arccos(x - 2) \) is \( 0 \le y \le \pi \). Hence:
\[ 0 \le \arccos(x - 2) \le \pi \]
Multiply all terms by \( -1 \) (which reverses inequality signs):
\[ 0 \ge -\arccos(x - 2) \ge -\pi \]
Rewriting gives the range of \( y = -\arccos(x - 2) \) as the interval \( [-\pi, 0] \).
Find the domain and range of \( y = -2 \arccos(-2x + 1) \)
1. Domain: Impose the condition on the argument \( (-2x + 1) \):
\[ -1 \le (-2x + 1) \le 1 \]
Solve to obtain domain: \( 0 \le x \le 1 \)
2. Range: The range of \( \arccos(-2x + 1) \) is \( 0 \le y \le \pi \). Hence:
\[ 0 \le \arccos(-2x + 1) \le \pi \]
Multiply by \( -2 \) (reversing inequalities):
\[ 0 \ge -2\arccos(-2x + 1) \ge -2\pi \]
Thus the range is \( [-2\pi, 0] \).
Find the domain and range of \( y = 2 \arccos(2x) + \frac{\pi}{2} \)
1. Domain: Condition on \( 2x \):
\[ -1 \le 2x \le 1 \]
Solve: \( -\frac{1}{2} \le x \le \frac{1}{2} \)
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 \( \frac{\pi}{2} \): \( \frac{\pi}{2} \le 2\arccos(2x) + \frac{\pi}{2} \le \frac{5\pi}{2} \)
Thus the range is \( \left[ \frac{\pi}{2}, \frac{5\pi}{2} \right] \).