Question
Periodic Function Flexibility Quick Check
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Question
Determine the period of the function f(x)=2sin(3x+π3)+5.(1 point)
Responses
2π3
Start Fraction 2 pi over 3 End Fraction
2π
2 pi
π9
Start Fraction pi over 9 End Fraction
π
pi
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Answers
GPT-4o mini
To determine the period of the function \( f(x) = 2\sin(3x + \frac{\pi}{3}) + 5 \), we focus on the sine part of the function.
The general form for the sine function is \( \sin(kx) \), where \( k \) affects the period. The period \( P \) of the sine function is given by the formula:
\[
P = \frac{2\pi}{k}
\]
In your function, the expression inside the sine function is \( 3x + \frac{\pi}{3} \). Here, \( k = 3 \).
Now we can substitute \( k \) into our period formula:
\[
P = \frac{2\pi}{3}
\]
Thus, the period of the function \( f(x) = 2\sin(3x + \frac{\pi}{3}) + 5 \) is \( \frac{2\pi}{3} \).
The correct answer is:
**\( \frac{2\pi}{3} \)**
The general form for the sine function is \( \sin(kx) \), where \( k \) affects the period. The period \( P \) of the sine function is given by the formula:
\[
P = \frac{2\pi}{k}
\]
In your function, the expression inside the sine function is \( 3x + \frac{\pi}{3} \). Here, \( k = 3 \).
Now we can substitute \( k \) into our period formula:
\[
P = \frac{2\pi}{3}
\]
Thus, the period of the function \( f(x) = 2\sin(3x + \frac{\pi}{3}) + 5 \) is \( \frac{2\pi}{3} \).
The correct answer is:
**\( \frac{2\pi}{3} \)**