Questions on how to find domain and range of arccosine functions.

**Theorems**

1.
**y = arccos x is equivalent to cos y = x**

with **-1 ≤ x ≤ 1 and 0 ≤ y ≤ pi **

**Question 1:** Find the domain and range of y = arccos(x + 1)

**Solution to question 1:**

1. Domain: To find the domain of the above function, we need to impose a condition on the argument (x + ) according to the domain of arccos(x) which is -1 ≤ x ≤ 1 . Hence

-1 ≤ (x + 1) ≤ 1

solve to obtain domain as: - 2 ≤ x ≤ 0

which as expected means that the graph of y = arccos(x + 1) is that of y = arccos(x) shifted one unit to the left.

2. Range: A shift to the left does not affect the range. Hence the range of y = arccos(x + 1) is the same as the range of arcsin(x) which is 0 ≤ y ≤ pi

**Question 2:** Find the domain and range of y = - arccos(x - 2)

**Solution to question 2:**

1. Domain: To find the domain of the above function, we need to impose a condition on the argument (x - 2) according to the domain of arccos(x) which is -1 ≤ x ≤ 1 . Hence

-1 ≤ (x - 2) ≤ 1

solve to obatain domain as: 1 ≤ x ≤ 3

which as expected means that the graph of y = arccos(x - 2) is that of y = arcsin(x) shifted two units to the right.

2. Range: The range of arccos(x - 2) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write

0 ≤ arcsin(x + 2) ≤ pi

We now multiply all terms of the above inequality by - 1 and invert the inequality symbols

0 ≥ - arcsin(x + 2) ≥ pi

which gives the range of y = - arccos(x + 2) as the interval [0 , pi]

**Question 3:** Find the domain and range of y = -2 arccos(- 2x + 1)

**Solution to question 3:**

1. Domain: To find the domain, we need to impose the following condition

-1 ≤ (- 2 x + 1) ≤ 1

solve to obtain domain as: 0 ≤ x ≤ 1

2. Range: The range of arccos(- 2x + 1) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write

0 ≤ arccos(- 2x + 1) ≤ pi

We now multiply all terms of the above inequality by - 2 and invert the inequality symbols

0 ≥ - 2 arccos(3x - 1) ≥ - 2 pi

which gives the range of y = - 2 arccos(- 2x + 1) as the interval [- 2 pi]

**Question 4:** Find the domain and range of y = 2 arccos( 2 x ) + pi/2

**Solution to question 4:**

1. Domain: To find the domain, we need to impose the following condition

-1 ≤ 2x ≤ 1

solve to obtain domain as: - 1 / 2 ≤ x ≤ 1 / 2

2. Range: The range of arccos(2x) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write

0 ≤ arccos(2x) ≤ pi

We now multiply all terms of the above inequality by 2

0 ≥ 2 arccos(2x) ≥ 2 pi

We now subtract pi/2 to all terms of the above inequality.

pi / 2 ≥ 2 arccos(2x) + pi / 2 ≥ 5 pi / 2

which gives the range of 2 arccos(2x) + pi / 2 as the interval [ pi / 2 , 5 pi / 2]