Consider the function $f(x) = \sin(x)$ in the interval $x \in [\pi/4, 7\pi/4]$. The number and location(s) of the local minima of this function are

Given $f(x) = \sin(x)$ on the closed interval $[\pi/4, 7\pi/4]$:

1. Interior critical points:

   $$f^{\prime}(x) = \cos(x) = 0$$

   In the interior $(\pi/4, 7\pi/4)$, the solutions are $x = \pi/2$ and $x = 3\pi/2$.

   • At $x = \pi/2$, $f^{\prime\prime}(x) = -\sin(\pi/2) = -1 < 0$, which is a local maximum.

   • At $x = 3\pi/2$, $f^{\prime\prime}(x) = -\sin(3\pi/2) = 1 > 0$, which is a local minimum.

2. Boundary/Endpoint behavior:

   • At the left endpoint $x = \pi/4$, $f^{\prime}(\pi/4) = \cos(\pi/4) = 1/\sqrt{2} > 0$. Since the derivative is strictly positive, $f(x)$ is increasing as $x$ moves to the right of $\pi/4$. Thus, $f(\pi/4) \le f(x)$ for all $x$ in a right-neighborhood of $\pi/4$, making $x = \pi/4$ a local minimum.

   • At the right endpoint $x = 7\pi/4$, $f^{\prime}(7\pi/4) = \cos(7\pi/4) = 1/\sqrt{2} > 0$, so $f(x)$ increases as $x \to 7\pi/4$ from the left, making $x = 7\pi/4$ a local maximum.

Therefore, there are two local minima, located at $x = \pi/4$ and $x = 3\pi/2$.

This matches option (D).