consider the function ( y=cos (x) ). which change would increase the period by a factor of 3?\nmultiply (…

Explanation:

Step1: Recall the period formula for (y = A\cos(Bx))

The period of the function (y = A\cos(Bx)) is (T=\frac{2\pi}{|B|}). For (y = \cos(x)), (B = 1) and (T = 2\pi).

Step2: Analyze each option

• Option 1: (y=\frac{1}{3}\cos(x)) Here (B = 1), so the period (T=\frac{2\pi}{1}=2\pi). Multiplying (\cos(x)) by (\frac{1}{3}) changes the amplitude (vertical - stretch/shrink), not the period.
• Option 2: (y = 3\cos(x)) Here (B = 1), so the period (T=\frac{2\pi}{1}=2\pi). Multiplying (\cos(x)) by (3) changes the amplitude (vertical - stretch/shrink), not the period.
• Option 3: (y=\cos(\frac{1}{3}x)) Using the period formula (T=\frac{2\pi}{|B|}), with (B=\frac{1}{3}), we have (T=\frac{2\pi}{\frac{1}{3}}=6\pi). The original period of (y = \cos(x)) is (2\pi), and (6\pi=3\times2\pi).
• Option 4: (y=\cos(3x)) Using the period formula (T=\frac{2\pi}{|B|}), with (B = 3), we have (T=\frac{2\pi}{3}). This is a decrease in the period.

Answer:

C. Multiply (x) by (\frac{1}{3})