Q:

1. which function is shown on the graph?f(x)=1/2cosxf(x)=βˆ’1/2sinxf(x)=βˆ’1/2cosxf(x)=1/2sinx2. (picture)3.(picture)4.(which equation represents the function on the graph?5. what is the period of the funtion f(x)=cos2x?

Accepted Solution

A:
Problem 1

Answer: choice C) f(x) = (-1/2)cos(x)

We can rule out anything with sine in it since the graph does not go through the origin. We can rule out choice A as well since plugging x = 0 into f(x) leads to a positive result, when instead we want a negative value. So the only thing left is choice C

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Problem 2

See the attached image

Point A is approximately (1.57, 0) which is exactly (pi/2, 0)
Point B is approximately (3.14, -2) which is exactly (pi, -2)

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Problem 3

Answer: 1/pi

Period = pi since the graph repeats itself every pi units
Frequency = 1/PeriodΒ 
Frequency = 1/piΒ 

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Problem 4

Answer: Choice A) f(x) = cos(2x)

This is a cosine function as it doesn't go through the origin. The period is T = pi, so b = 2pi/T = 2pi/pi = 2 is the coefficient of the inner x term. Therefore the function is f(x) = cos(2x)

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