Question
Periodic Functions Online Practice Part 2
Complete this assessment to review what you’ve learned. It will not count toward your grade.
10 of 1110 of 11 Items
Question
Determine the period of the function f(x)=2cos(16x+π3)
.(1 point)
Responses
12π
12 pi
2π
2 pi
π3
Start Fraction pi over 3 End Fraction
−π3
negative Start Fraction pi over 3 End Fraction
Remaining Attempts : 3
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Answers
GPT-4o mini
To determine the period of the function \( f(x) = 2\cos(16x + \frac{\pi}{3}) \), we can use the formula for the period of a cosine function, which is given by:
\[
\text{Period} = \frac{2\pi}{|b|}
\]
where \( b \) is the coefficient of \( x \) in the argument of the cosine function \( \cos(bx + c) \).
In the function \( f(x) = 2\cos(16x + \frac{\pi}{3}) \), the value of \( b \) is 16.
Now we can calculate the period:
\[
\text{Period} = \frac{2\pi}{|16|} = \frac{2\pi}{16} = \frac{\pi}{8}
\]
Since this specific option isn't given in your responses, let's confirm our calculations and take another look at the provided options:
- \( 12\pi \)
- \( 2\pi \)
- \( \frac{\pi}{3} \)
- \( -\frac{\pi}{3} \)
None of these actually corresponds to \( \frac{\pi}{8} \). It seems there might be a misunderstanding in the provided question or options. However, based on our calculations, the period we've derived is \( \frac{\pi}{8} \), which isn't present in the response options provided. Thus, we can conclude with our computations, but please double-check the question context or possible options available.
\[
\text{Period} = \frac{2\pi}{|b|}
\]
where \( b \) is the coefficient of \( x \) in the argument of the cosine function \( \cos(bx + c) \).
In the function \( f(x) = 2\cos(16x + \frac{\pi}{3}) \), the value of \( b \) is 16.
Now we can calculate the period:
\[
\text{Period} = \frac{2\pi}{|16|} = \frac{2\pi}{16} = \frac{\pi}{8}
\]
Since this specific option isn't given in your responses, let's confirm our calculations and take another look at the provided options:
- \( 12\pi \)
- \( 2\pi \)
- \( \frac{\pi}{3} \)
- \( -\frac{\pi}{3} \)
None of these actually corresponds to \( \frac{\pi}{8} \). It seems there might be a misunderstanding in the provided question or options. However, based on our calculations, the period we've derived is \( \frac{\pi}{8} \), which isn't present in the response options provided. Thus, we can conclude with our computations, but please double-check the question context or possible options available.