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Question

Question: How do you graph the given function \(y=3\cos 2\pi x\) and include two full periods?...

How do you graph the given function y=3cos2πxy=3\cos 2\pi x and include two full periods?

Explanation

Solution

We start solving the problem by finding the period of the given function by using the fact that the period of the function acosbxa\cos bx is defined as 2πb\dfrac{2\pi }{\left| b \right|}. We then find the range of the given function by using the fact that the range of the function acosbxa\cos bx lies in the interval [a,a]\left[ -a,a \right]. We then find the values of x at which we get maximum, minimum and 0. We then plot these points to get the required graph of the given function.

Complete step by step answer:
According to the problem, we are asked to graph the given function y=3cos2πxy=3\cos 2\pi x by including two full periods.
We have given the function y=3cos2πxy=3\cos 2\pi x ---(1).
Let us first find the period of the given function. We know that the period of the function acosbxa\cos bx is defined as 2πb\dfrac{2\pi }{\left| b \right|}. Let us use this result in equation (1).
So, we get the period of the function y=3cos2πxy=3\cos 2\pi x as 2π2π=1\dfrac{2\pi }{2\pi }=1.
Now, we need to plot the given function y=3cos2πxy=3\cos 2\pi x for two full periods, which is [0,2]\left[ 0,2 \right].
We know that the range of the function acosbxa\cos bx lies in the interval [a,a]\left[ -a,a \right]. So, the range of the given function y=3cos2πxy=3\cos 2\pi x is [3,3]\left[ -3,3 \right].
We know that the maximum value of acosbxa\cos bx occurs at bx=2nπbx=2n\pi , nZn\in Z. So, the maximum value of y=3cos2πxy=3\cos 2\pi x i.e., y=3y=3 occurs at 2πx=2nπx=n2\pi x=2n\pi \Leftrightarrow x=n, nZn\in Z ---(1).
We know that the minimum value of acosbxa\cos bx occurs at bx=(2n+1)πbx=\left( 2n+1 \right)\pi , nZn\in Z. So, the maximum value of y=3cos2πxy=3\cos 2\pi x i.e., y=3y=-3 occurs at 2πx=(2n+1)πx=n+122\pi x=\left( 2n+1 \right)\pi \Leftrightarrow x=n+\dfrac{1}{2}, nZn\in Z ---(2).
We know that the acosbx=0a\cos bx=0 occurs at bx=(2n+1)π2bx=\left( 2n+1 \right)\dfrac{\pi }{2}, nZn\in Z. So, the maximum value of y=3cos2πxy=3\cos 2\pi x i.e., y=0y=0 occurs at 2πx=(2n+1)π2x=2n+142\pi x=\left( 2n+1 \right)\dfrac{\pi }{2}\Leftrightarrow x=\dfrac{2n+1}{4}, nZn\in Z ---(3).
Now, let us plot the given function by using the results obtained from equations (1), (2) and (3).

Note:
We should not confuse while finding the general solution for the maximum, minimum values of the given function. We should keep in mind that the period will always be greater than 0 while solving this type of problem. We should not make calculation mistakes while solving this type of problem. Similarly, we can expect problems graphing the given function y=tan2πxy=\tan 2\pi x and include four full periods.