Question
Question: How do you graph \(f(x) = 1 - \cos 3x\)?...
How do you graph f(x)=1−cos3x?
Solution
This problem deals with finding the graph of the given function. The function involves a trigonometric function which is a cosine trigonometric function. The general equation or a general function of a cosine function is denoted by Acos(B(x−C))+D. Where this particular function has a period, amplitude and phase shift. The period of a periodic function is the interval between two matching points on the graph. The period of cosine function is 2π.
Complete step-by-step answer:
Any function is denoted in such a way, where f(x)=Acos(B(x−C))+D
⇒f(x)=Acos(B(x−C))+D
Here A is the amplitude of the function, P is the period of the function, C is the phase shift and D is the vertical shift of the given function.
⇒P=B2π
Here on comparing the given function f(x)=1−cos3x with the standard form of cosine function which is f(x)=Acos(B(x−C))+D
⇒f(x)=1−cos3x
Here there are two things to be observed which are the amplitude of the function is -1 and the vertical shift is 1. As the amplitude is negative, the graph will be on the side of the negative vertical axis.
Here the period is by P=B2π, as given below, here B is 3.
⇒P=32π
There is no phase shift. So the graph of f(x)=1−cos3x, is given below:
Final Answer: The amplitude, period, phase shift and vertical shift for f(x)=1−cos3x are -1, 32π , 0 and 1 respectively.
Note:
Please note that the fundamental period of a function is the period of the function which are of the form, f(x+k)=f(x) and f(x)=f(x+k), then k is called the period of the function and the function f is called a periodic function. In other words, it is the distance along the x-axis that the function has to travel before it starts to repeat its pattern. The basic sine and cosine functions have a period of 2π, while tangent has a period of π.