Question

assignment 8.1: graphs of the sine and cosine functions
score: 17/21 answered: 6/10
question 7
sketch a graph of the function ( f(x)=5 cos left(\frac{1}{3} x
ight) )
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Explanation:

Step1: Find the amplitude

The general form of a cosine function is \(y = A\cos(Bx)\). For the function \(f(x)=5\cos(\frac{1}{3}x)\), the amplitude \(|A|\). Here \(A = 5\), so the amplitude is \(5\). This means the graph oscillates between \(y = 5\) and \(y=-5\).

Step2: Find the period

The period of the function \(y=\cos(Bx)\) is given by \(T=\frac{2\pi}{|B|}\). For \(f(x)=5\cos(\frac{1}{3}x)\), \(B=\frac{1}{3}\). Then \(T=\frac{2\pi}{\frac{1}{3}}=6\pi\).

Step3: Plot key points

• When \(x = 0\), \(f(0)=5\cos(0)=5\)
• When \(x=\frac{3\pi}{2}\), \(f(\frac{3\pi}{2})=5\cos(\frac{1}{3}\times\frac{3\pi}{2})=5\cos(\frac{\pi}{2}) = 0\)
• When \(x = 3\pi\), \(f(3\pi)=5\cos(\frac{1}{3}\times3\pi)=5\cos(\pi)=-5\)
• When \(x=\frac{9\pi}{2}\), \(f(\frac{9\pi}{2})=5\cos(\frac{1}{3}\times\frac{9\pi}{2})=5\cos(\frac{3\pi}{2}) = 0\)
• When \(x = 6\pi\), \(f(6\pi)=5\cos(\frac{1}{3}\times6\pi)=5\cos(2\pi)=5\)

Connect these key points with a smooth curve that has an amplitude of \(5\) and a period of \(6\pi\) to sketch the graph of \(y = 5\cos(\frac{1}{3}x)\).

Answer:

Sketch a cosine - type curve with amplitude \(5\) (reaching up to \(y = 5\) and down to \(y=-5\)) and period \(6\pi\). Mark key points \((0,5)\), \((\frac{3\pi}{2},0)\), \((3\pi,-5)\), \((\frac{9\pi}{2},0)\), \((6\pi,5)\) and connect them with a smooth curve.