(1 point) Find the inflection points and local extrema of $f(x)=12 \cos x-6 \sqrt{2} x$ with domain $-\pi \leq x \leq 3 \pi / 2$. Then find the intervals on which $f(x)$ is concave up and concave down. Inflection points occur at (Enter a comma-separated list of $x$-coordinates. If there are no inflection points, enter "none".) Local maxima occur at (Enter a comma-separated list of $x$-coordinates. If there are no local maxima, enter "none".) Local minima occur at (Enter a comma-separated list of $x$-coordinates. If there are no local minima, enter "none".) $f(x)$ is concave up on $f(x)$ is concave down on (Enter your answers using interval notation.)

Step 1 ：To find the inflection points and local extrema of the function \( f(x)=12 \cos x-6 \sqrt{2} x \), we need to find the first and second derivatives of the function.

Step 2 ：The first derivative is \( f'(x) = -12 \sin x - 6 \sqrt{2} \).

Step 3 ：The critical points are found by setting the first derivative equal to zero: \( -12 \sin x - 6 \sqrt{2} = 0 \).

Step 4 ：Solving for \( x \), we get the critical points at \( x = -\frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Step 5 ：The second derivative is \( f''(x) = -12 \cos x \).

Step 6 ：Inflection points occur where the second derivative changes sign. Setting \( f''(x) = 0 \), we find inflection points at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Step 7 ：Using the first and second derivatives, we can determine that local maxima occur at \( x = -\frac{\pi}{4} \) and local minima occur at \( x = \frac{5\pi}{4} \).

Step 8 ：The function is concave up on the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) and concave down on the intervals \( (-\pi, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, \frac{3\pi}{2}) \).

Step 9 ：Final Answer: Inflection points occur at \(\boxed{\frac{\pi}{2}, \frac{3\pi}{2}}\), Local maxima occur at \(\boxed{-\frac{\pi}{4}}\), Local minima occur at \(\boxed{\frac{5\pi}{4}}\), \( f(x) \) is concave up on \(\boxed{(\frac{\pi}{2}, \frac{3\pi}{2})}\), \( f(x) \) is concave down on \(\boxed{(-\pi, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, \frac{3\pi}{2})}\)