Identify the horizontal middle of a sample period for the function:

Question

f (t) = 4sec(3t − 6π)

 

a. x=4pi/3

b. x=5pi/2

c. x=pi

d. x=3+pi

Answer 1

1. Horizontal Middle of a Sample Period  The horizontal middle of a sample period for the function \(f(t)=4\sec (3t-6\pi )\) is determined by finding the phase shift and the period.  Step-by-step Solution  Determine the phase shift:  The phase shift is found by setting the argument of the secant function to zero and solving for \(t\). \(3t-6\pi =0\) \(3t=6\pi \) \(t=\frac{6\pi }{3}\) \(t=2\pi \) Therefore, the phase shift is \(2\pi \). This represents the beginning of a sample period.  Calculate the period: The period \(T\) of a secant function in the form \(a\sec (bt-c)\) is given by \(T=\frac{2\pi }{|b|}\). In this function, \(b=3\). \(T=\frac{2\pi }{|3|}\) \(T=\frac{2\pi }{3}\)Identify the end of the sample period:

The end of the sample period is found by adding the period to the phase shift.

End of period

=Phase Shift+Periodequals Phase Shift plus Period

=Phase Shift+Period

End of period

=2π+2π3equals 2 pi plus the fraction with numerator 2 pi and denominator 3 end-fraction

=2𝜋+2𝜋3

End of period

=6π3+2π3equals the fraction with numerator 6 pi and denominator 3 end-fraction plus the fraction with numerator 2 pi and denominator 3 end-fraction

=6𝜋3+2𝜋3

End of period

=8π3equals the fraction with numerator 8 pi and denominator 3 end-fraction

=8𝜋3

Therefore, the end of the sample period is 8π3the fraction with numerator 8 pi and denominator 3 end-fraction

8𝜋3

. 

1. Determine the horizontal middle of the sample period:

The horizontal middle of the sample period is the average of the beginning and the end of the period.

Horizontal Middle

=Beginning of Period+End of Period2equals the fraction with numerator Beginning of Period plus End of Period and denominator 2 end-fraction

=Beginning of Period+End of Period2. 

1. Determine the horizontal middle of the sample period:

The horizontal middle of the sample period is the average of the beginning and the end of the period.

Horizontal Middle

=Beginning of Period+End of Period2equals the fraction with numerator Beginning of Period plus End of Period and denominator 2 end-fraction

=Beginning of Period+End of Period2

Horizontal Middle

=2π+8π32equals the fraction with numerator 2 pi plus the fraction with numerator 8 pi and denominator 3 end-fraction and denominator 2 end-fraction

=2𝜋+8𝜋32

Horizontal Middle

=6π3+8π32equals the fraction with numerator the fraction with numerator 6 pi and denominator 3 end-fraction plus the fraction with numerator 8 pi and denominator 3 end-fraction and denominator 2 end-fraction

=6𝜋3+8𝜋32

Horizontal Middle

=14π32equals the fraction with numerator the fraction with numerator 14 pi and denominator 3 end-fraction and denominator 2 end-fraction

=14𝜋32

Horizontal Middle

=14π3×2equals the fraction with numerator 14 pi and denominator 3 cross 2 end-fraction

=14𝜋3×2

Horizontal Middle

=14π6equals the fraction with numerator 14 pi and denominator 6 end-fraction

=14𝜋6

Horizontal Middle

=7π3equals the fraction with numerator 7 pi and denominator 3 end-fraction

=7𝜋3

Therefore, the horizontal middle of the sample period is 7π3the fraction with numerator 7 pi and denominator 3 end-fraction

7𝜋3

. 

Final Answer 

The horizontal middle of a sample period for the function

f(t)=4sec(3t−6π)f of t equals 4 secant open paren 3 t minus 6 pi close paren

𝑓(𝑡)=4sec(3𝑡−6𝜋)

is 7π3the fraction with numerator 7 pi and denominator 3 end-fraction

7𝜋3

.