Consider the function $f(x)=x^{4}-32 x^{2}+4,-3 \leq x \leq 9$. Use the derivatives to algebrically answer the question:
This function has an absolute minimum value equal to and an absolute maximum value equal to

Step 1 ：Consider the function \(f(x)=x^{4}-32 x^{2}+4\), for \(-3 \leq x \leq 9\).

Step 2 ：First, we need to find the derivative of the function \(f(x)=x^{4}-32 x^{2}+4\).

Step 3 ：The derivative of the function is \(f'(x) = 4x^{3}-64x\).

Step 4 ：Critical points occur where the derivative of the function is zero or undefined. So, we set the derivative equal to zero and solve for x to find the critical points.

Step 5 ：The critical points of the function are \(-4, 0, 4\).

Step 6 ：We then evaluate the function at these points and at the endpoints of the interval \([-3, 9]\) to find the absolute minimum and maximum values.

Step 7 ：The values of the function at these points are \(-252, 4, -252, -203, 3973\).

Step 8 ：The absolute minimum value of the function is \(-252\) and the absolute maximum value is \(3973\).

Step 9 ：Final Answer: The absolute minimum value of the function is \(\boxed{-252}\) and the absolute maximum value is \(\boxed{3973}\).