# Find Domain and Range of Arccosine Functions

 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]