(1 point) Find the inflection points and local extrema of $f(x)=12\cos x-6\sqrt{2}x$ with domain $-\pi \le x\le 3\pi /2$. 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 $f^{\prime }(x)=-12\sin x-6\sqrt{2}$.

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 $\boxed{{\displaystyle \frac{\pi }{2},\frac{3\pi }{2}}}$, Local maxima occur at $\boxed{{\displaystyle -\frac{\pi }{4}}}$, Local minima occur at $\boxed{{\displaystyle \frac{5\pi }{4}}}$, $f(x)$ is concave up on $\boxed{{\displaystyle (\frac{\pi }{2},\frac{3\pi }{2})}}$, $f(x)$ is concave down on $\boxed{{\displaystyle (-\pi ,\frac{\pi }{2})\cup (\frac{3\pi }{2},\frac{3\pi }{2})}}$