Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
End-of-Course Assessment

Exercise 82 Page 971

Practice makes perfect
a We are asked to find the period and amplitude of the graph of f(x).
f(x)= 2 cos( 4x) We can do this without graphing it. To do so, let's recall the general form of a cosine function. y= a cos( b θ) Here, | a| is the amplitude, b is the number of cycles in the interval from 0 to 2π, and 2π b is the period of the function. Let's first consider our amplitude by substituting a= 2. Amplitude: | 2|= 2 Next, we will calculate our period by substituting b= 4.
Period=2π/b
Period= 2π/4
Period= π/2
We found that the amplitude of the given function is 2 and the period is π2.
b Now we will graph f(x) over two periods. Let's first sketch only one cycle. To do so we will use the information that we found in Part A.
Amplitude: 2 Period: π/2 Now, to plan our sketch of the graph we will divide the period into fourths to plot the five-point pattern.

π/2 ÷ 4=π/8 Let's make a table of values to show these five points. We will start with 0.

2cos(4x) f(x)=2 cos(4x) Point
x=0 2 cos(4* 0) 2 cos(0)=2 ( 0,2)
x=π/8 2 cos (4* π/8) 2 cos (π/2)= ( π/8, )
x=π/4 2 cos (4* π/4 ) 2cos(π)=-2 ( π/4,-2 )
x=3π/8 2 cos (4* 3π/8) 2 cos (3π/2)= ( 3π/8, )
x=π/2 2 cos (4* π/2) 2cos(2π)=2 ( π/2,2)

Notice that the maximums will occur at 2 and the minimums will occur at -2 because the amplitude of the graph is 2. Let's now see these points on the coordinate plane!

Finally, we can connect the points to graph one cycle of the function.

Great! Now, we will also continue the pattern for one more cycle to have it over two periods.


c This time we will find the value which gives 0.5. To do so, we will set f(x) to 0.5 and then solve it for x.
0.5= 2cos( 4x) ⇒ 0.25=cos( 4x) Now, we will use the inverse cosine function to isolate 4x. 0.25=cos( 4x) ⇒ cos^(-1)(0.25)= 4x Let's solve the inverse function for x.
cos^(-1)(0.25)=4x
cos^(-1)(0.25)/4=x
x=cos^(-1)(0.25)/4
x=0.329529 ...
x ≈ 0.33
Remember that a cosine function takes the same value for θ and - θ in the interval 0 ≤ θ ≤ 2π. With this in mind, we can consider -0.33 as the other solution that gives 0.5. x ≈ 0.33 or x ≈ - 0.33 Finally, we need to expand our solutions by adding the multiple of our period π2. x ≈ 0.33+ π/2k or x ≈ - 0.33+ π/2k where k is an integer