Question

the sinusoidal function ( y = g(x) ) has a period of ( \frac{6pi}{5} ) and a minimum value at ( x = -\frac{3pi}{5} ). which of the following statements with reason is true?

a. the first maximum value for ( x > -\frac{3pi}{5} ) occurs at ( x = -\frac{pi}{5} ), because the smallest interval of input values between the maximum and minimum output values is ( pi ).

b. the first maximum value for ( x > -\frac{3pi}{5} ) occurs at ( x = -\frac{pi}{4} ), because the smallest interval of input values between the maximum and minimum output values is ( \frac{1}{2} ) of the period of the sinusoidal function.

c. the first maximum value for ( x > -\frac{3pi}{5} ) occurs at ( x = \frac{pi}{5} ), because the smallest interval of input values between the maximum and minimum output values is ( 2pi ).

d. the first maximum value for ( x > -\frac{3pi}{5} ) occurs at ( x = pi ), because the smallest interval of input values between the maximum and minimum output values is the period of the sinusoidal function.

Explanation:

Step1: Recall sinusoidal properties

For a sinusoidal function, the horizontal distance between a minimum and the next maximum is $\frac{1}{2}$ of the period. Given period $P=\frac{5\pi}{2}$, so this distance is $\frac{1}{2} \times \frac{5\pi}{2} = \frac{5\pi}{4}$.

Step2: Calculate first maximum position

The minimum is at $x=-\frac{3\pi}{2}$. Add the distance from min to max:

$$ -\frac{3\pi}{2} + \frac{5\pi}{4} = -\frac{6\pi}{4} + \frac{5\pi}{4} = -\frac{\pi}{4} $$

Step3: Verify interval claim

The interval between min and max is $\frac{1}{2}$ of the period, which matches option B's reasoning.

Answer:

B. The first maximum value for $x > -\frac{3\pi}{2}$ occurs at $x = -\frac{\pi}{4}$, because the smallest interval of input values between the maximum and minimum output values is $\frac{1}{2}$ of the period of the sinusoidal function.