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 geq 0 ).

( f(x)=4 sin left(\frac{1}{3} x
ight) )

Explanation:

Step1: Find the amplitude

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For the function \(f(x)=4\sin(\frac{1}{3}x)\), the amplitude \(|A|\). Here \(A = 4\), so the amplitude is \(4\).

Step2: Find the period

The period of a sine function \(y=\sin(Bx)\) is given by \(T=\frac{2\pi}{|B|}\). For \(y = 4\sin(\frac{1}{3}x)\), \(B=\frac{1}{3}\). Then \(T=\frac{2\pi}{\frac{1}{3}}=6\pi\).

Step3: Find the midline

For the function \(y = A\sin(Bx - C)+D\), \(D = 0\) (since \(f(x)=4\sin(\frac{1}{3}x)+0\)). So the mid - line is \(y = 0\).

Step4: Find the maximum and minimum values and corresponding \(x\) - values

The maximum value of \(y=\sin t\) is \(1\) and the minimum is \(- 1\). For \(y = 4\sin(\frac{1}{3}x)\):

• When \(\sin(\frac{1}{3}x)=1\), \(\frac{1}{3}x=\frac{\pi}{2}+2k\pi\) (\(k = 0\) for the first non - negative \(x\) in one period). Solving for \(x\), we get \(x=\frac{3\pi}{2}\), and \(y = 4\).
• When \(\sin(\frac{1}{3}x)=-1\), \(\frac{1}{3}x=\frac{3\pi}{2}+2k\pi\) (\(k = 0\) for the first non - negative \(x\) in one period). Solving for \(x\), we get \(x=\frac{9\pi}{2}\), and \(y=-4\).

Answer:

Amplitude: \(4\), Period: \(6\pi\), Midline: \(y = 0\), Maximum value \(y = 4\) at \(x=\frac{3\pi}{2}\), Minimum value \(y=-4\) at \(x=\frac{9\pi}{2}\)