Domain range and period

Question

a) what is the domain, range, and periodic y=csc x? 

b) what is the relative minimum in the interval 0<_x<_ pi

c) what is the relative maximum ion the interval pi<_x<_2pi 

 

(<_ is less that or equal to)

Answer 1

Remember that cosecant is the inverse of sine. In other words, csc(x) = 1/sin(x).

 

 

 

Since we can't ever divide by zero in math, the domain of csc(x) is every real number except whenever sin(x) = 0. When does that happen? Think about your unit circle or the graph of sin(x). Sin(x) = 0 whenever x = 0, π, 2π, 3π, .... So, x can't be a multiple of π. That means the domain of csc(x) is all real numbers except πn, where n is any integer.

 

 

 

Since csc(x) = 1/sin(x), we'll need to think about the possible values of sin(x) to figure out csc(x)'s range. Sin(x) has values between 1 and -1. That means that csc(x) will have no number whose absolute value is greater than 1 in its denominator. 1 divided by any number whose absolute value is less than 1 is greater than 1 (like 1/(1/2) = 2). We will never have a value for csc(x) between -1 and 1, since that would require 1 divided by a number whose absolute value is greater than 1 (like 1/2). We can, however, have 1/1 = 1 and 1/-1 = -1. So, the range of csc(x) is (-infinity, -1]∪[1, infinity).

 

 

 

Since csc(x) is the inverse of sin(x), the period of csc(x) is 2π, just like the period of sin(x).

 

 

 

Between 0 and π, the lowest value csc(x) will have occurs when sin(x) is at its greatest value. Sin(π/2) = 1, so csc(π/2) = 1/sin(π/2) = 1/1 = 1. The relative minimum is therefore (π/2, 1).

 

 

 

Between π and 2π, the highest value csc(x) will have occurs when sin(x) is at its lowest value. Sin(3π/2) = -1, so csc(3π/2) = 1/sin(3π/2) = 1/-1 = -1. The relative maximum is therefore (3π/2, -1).