Question

graph the function and state the amplitude, period, and midline. state the maximum and minimum y - values and the corresponding x - values on one period for x ≥ 0 and x < the period.

f(x)=3cos(3x)

clear all draw:

amplitude:

period:

Explanation:

Step1: Find the amplitude

For the function \(y = A\cos(Bx)\), the amplitude is \(|A|\).
For \(f(x)=3\cos(3x)\), \(A = 3\), so the amplitude \(a=\vert3\vert = 3\).

Step2: Find the period

The period of the function \(y=\cos(Bx)\) is \(T=\frac{2\pi}{|B|}\).
For \(f(x)=3\cos(3x)\), \(B = 3\), so the period \(T=\frac{2\pi}{3}\).

Step3: Find the mid - line

The mid - line of the function \(y = A\cos(Bx)\) is \(y = 0\) (since there is no vertical shift \(D\) in the general form \(y=A\cos(Bx - C)+D\)).

Step4: Find the maximum and minimum values

The maximum value of \(y = A\cos(Bx)\) occurs when \(\cos(Bx)=1\).
If \(\cos(3x)=1\), then \(3x = 2k\pi\) (\(k\in\mathbb{Z}\)). For \(x\geq0\) and \(x<\frac{2\pi}{3}\), when \(k = 0\), \(x = 0\), and \(y=3\times1=3\).
The minimum value of \(y = A\cos(Bx)\) occurs when \(\cos(Bx)=-1\).
If \(\cos(3x)=-1\), then \(3x=(2k + 1)\pi\) (\(k\in\mathbb{Z}\)). For \(x\geq0\) and \(x<\frac{2\pi}{3}\), when \(k = 0\), \(x=\frac{\pi}{3}\), and \(y=3\times(-1)=-3\).

Answer:

Amplitude: \(3\); Period: \(\frac{2\pi}{3}\); Mid - line: \(y = 0\); Maximum value: \(y = 3\) at \(x = 0\); Minimum value: \(y=-3\) at \(x=\frac{\pi}{3}\)