Find the absolute maximum and minimum values of each function over the indicated interval, and indicate the $x$-values at which they occur. \[ f(x)=3 x^{3}-3 x^{2}-3 x+5 ;[-1,0] \] The absolute maximum value is $\square$ at $x=$ (Use a comma to separate answers as needed. Type an integer or a fraction.)
Solution
Step 1 ：Find the derivative of the function \(f(x) = 3x^3 - 3x^2 - 3x + 5\).
Step 2 ：Set the derivative equal to zero to find the critical points.
Step 3 ：The critical points are \(-1/3\) and \(1\).
Step 4 ：Evaluate the function at the critical points and the endpoints of the interval \([-1, 0]\).
Step 5 ：The maximum value of the function on the interval \([-1, 0]\) is \(50/9\) and it occurs at \(x = -1/3\).
Step 6 ：The minimum value of the function on the interval \([-1, 0]\) is \(2\) and it occurs at \(x = 1\) and \(x = -1\).
Step 7 ：However, since \(x = 1\) is not within the interval \([-1, 0]\), the minimum value on this interval is \(2\) at \(x = -1\).
Step 8 ：Final Answer: The absolute maximum value is \(\boxed{\frac{50}{9}}\) at \(x=\boxed{-\frac{1}{3}}\).
