The general form of a sine function is: \[ y = \sin(kx) \] where the period of the sine function is given by: \[ {Period} = \frac{2\pi}{|k|} \] Here, \( k \) is the coefficient of \( x \) in the argument of the sine function. In our case, the function is \( \sin\left( \frac{\pi x}{4} \right) \), so \( k = \frac{\pi}{4} \).
Using the formula for the period, we get: \[ {Period} = \frac{2\pi}{\left|\frac{\pi}{4}\right|} = \frac{2\pi}{\frac{\pi}{4}} = 8 \] Thus, the period of the function \( \sin\left( \frac{\pi x}{4} \right) \) is \( 8 \). Thus, the correct answer is \( \boxed{8} \), corresponding to option (D).
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to: