### Instruction: Find the range of the function
\[f(x) = \left( \arccos \frac{x}{2} \right)^2 + \pi \arcsin \frac{x}{2} - \left( \arcsin \frac{x}{2} \right)^2 + \frac{\pi^2}{12} (x^2 + 6x + 8).\].### Response: I see that this function involves inverse trigonometric functions and some quadratic terms.
To find the range, I need to determine the minimum and maximum values that the function can take for any real value of x.
I also need to be careful about the domain of the function, since the inverse trigonometric functions have restricted domains.
Let me start by simplifying the function a bit.
I notice that the first term and the third term are both squared, and they have the same argument of x/2.
I wonder if I can use a trigonometric identity to relate them.
I recall that $\arccos y + \arcsin y = \frac{\pi}{2}$ for any y in the interval [-1, 1].
So I can write $\arccos \frac{x}{2} = \frac{\pi}{2} - \arcsin \frac{x}{2}$.
Then, squaring both sides, I get $\left( \arccos \frac{x}{2} \right)^2 = \left( \frac{\pi}{2} - \arcsin \frac{x}{2} \right)^2$.
Expanding the right-hand side, I get $\left( \arccos \frac{x}{2} \right)^2 = \frac{\pi^2}{4} - \pi \arcsin \frac{x}{2} + \left( \arcsin \frac{x}{2} \right)^2$.
Now I can substitute this expression into the original function and simplify.
I get $f(x) = \frac{\pi^2}{4} - \pi \arcsin \frac{x}{2} + \left( \arcsin \frac{x}{2} \right)^2 + \pi \arcsin \frac{x}{2} - \left( \arcsin \frac{x}{2} \right)^2 + \frac{\pi^2}{12} (x^2 + 6x + 8)$.
I can see that the second and fourth terms cancel out, and so do the third and fifth terms.
So the function simplifies to $f(x) = \frac{\pi^2}{4} + \frac{\pi^2}{12} (x^2 + 6x + 8)$.
This is a quadratic function of x, with a positive coefficient for the $x^2$ term.
That means it has a minimum value at the vertex of the parabola, and it increases without bound as x goes to infinity or negative infinity.
To find the vertex, I can use the formula $x = -b/2a$, where $a = \frac{\pi^2}{12}$ and $b = 6 \frac{\pi^2}{12}$.
So the vertex is at $x = -3$.
To find the minimum value of the function, I can plug in $x = -3$ and simplify.
I get $f(-3) = \frac{\pi^2}{4} + \frac{\pi^2}{12} ((-3)^2 + 6(-3) + 8)$.
This simplifies to $f(-3) = \frac{\pi^2}{4} - \frac{\pi^2}{12}$.
So the minimum value of the function is $\frac{\pi^2}{6}$.
The maximum value of the function is positive infinity, since the function increases without bound as x goes to infinity or negative infinity.
Therefore, the range of the function is the interval $[\frac{\pi^2}{6}, \infty)$.