To find the absolute maximum and minimum values of the function $f(x) = 2x^3 - x^2 - 4x + 6$ over the indicated interval, we need to follow these steps:

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Steps:

1. Find the derivative of $f(x)$: The derivative $f'(x)$ will help us locate the critical points. $f'(x) = 6x^2 - 2x - 4$

2. Set $f'(x) = 0$ to find critical points: Solve $6x^2 - 2x - 4 = 0$. This quadratic equation can be solved using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 6$, $b = -2$, $c = -4$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(6)(-4)}}{2(6)}$ $x = \frac{2 \pm \sqrt{4 + 96}}{12}$ $x = \frac{2 \pm \sqrt{100}}{12}$ $x = \frac{2 \pm 10}{12}$ $x = \frac{12}{12} = 1 \quad \text{or} \quad x = \frac{-8}{12} = -\frac{2}{3}$

3. Evaluate $f(x)$ at critical points and endpoints: The absolute maximum and minimum values of a continuous function over a closed interval occur either at critical points or endpoints.

   Suppose the interval is $[a, b]$. Evaluate $f(x)$ at:

   • Critical points $x = 1$ and $x = -\frac{2}{3}$,
   • Endpoints $x = a$ and $x = b$.

4. Compare the results: The largest value is the absolute maximum, and the smallest value is the absolute minimum.

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Let me calculate the exact values for you. Could you confirm the interval (e.g., $[-2, 2]$) or provide one?