Question 9

We are given $f(x) = \sin(x)$ on the closed interval $[\pi/4, 7\pi/4]$.

1. Interior critical points:
   Set the derivative to zero:

   $$f'(x) = \cos(x) = 0$$

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

   • At $x = \pi/2$: $f''(\pi/2) = -\sin(\pi/2) = -1 < 0 \implies$ local maximum.
   • At $x = 3\pi/2$: $f''(3\pi/2) = -\sin(3\pi/2) = -(-1) = 1 > 0 \implies$ local minimum.
     The function value is $f(3\pi/2) = -1$.

2. Endpoint analysis:
   For a function defined on a closed interval $[a, b]$, an endpoint is a local minimum if the function values in an adjacent one-sided neighborhood are greater than or equal to the endpoint value:

   • Left endpoint $x = \pi/4$:
     Since $f'(\pi/4) = \cos(\pi/4) = 1/\sqrt{2} > 0$, the function is strictly increasing immediately to the right of $\pi/4$. Thus, for all $x \in (\pi/4, \pi/4 + \delta)$, $f(x) > f(\pi/4)$. By definition, $x = \pi/4$ is a local minimum.
   • Right endpoint $x = 7\pi/4$:
     Since $f'(7\pi/4) = \cos(7\pi/4) = 1/\sqrt{2} > 0$, the function was increasing as it approached $7\pi/4$. Thus $f(x) < f(7\pi/4)$ for points immediately to the left, meaning $7\pi/4$ is a local maximum, not a local minimum.

Therefore, there are two local minima: one at $\pi/4$ (the left boundary) and one at $3\pi/2$ (the interior critical point).
This corresponds to option (D).