To find the absolute maximum value of the function $f(x) = 2x^2 - 2x - 2$ on the interval $[0, 6]$, follow these steps:

Step 1: Compute the derivative

The first step is to find the critical points by taking the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(2x^2 - 2x - 2) = 4x - 2$

Step 2: Find critical points

Set $f'(x) = 0$ to solve for $x$: $4x - 2 = 0 \quad \implies \quad x = \frac{2}{4} = \frac{1}{2}.$

Step 3: Evaluate $f(x)$ at the critical point and endpoints

We need to evaluate $f(x)$ at $x = 0$, $x = 6$, and $x = \frac{1}{2}$.

1. At $x = 0$: $f(0) = 2(0)^2 - 2(0) - 2 = -2$

2. At $x = \frac{1}{2}$: $f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - 2$ $f\left(\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) - 1 - 2 = \frac{1}{2} - 1 - 2 = -\frac{5}{2} = -2.5$

3. At $x = 6$: $f(6) = 2(6)^2 - 2(6) - 2 = 2(36) - 12 - 2 = 72 - 14 = 58$

Step 4: Determine the absolute maximum

The values of $f(x)$ are:

• $f(0) = -2$
• $f\left(\frac{1}{2}\right) = -2.5$
• $f(6) = 58$

The absolute maximum value of $f(x)$ on $[0, 6]$ is: $\boxed{58 \text{ at } x = 6.}$

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5 Related Questions:

1. How do you determine the absolute minimum value on a closed interval?
2. What happens if there are no critical points inside the interval?
3. How do you verify if a critical point is a maximum or minimum?
4. How do absolute extrema differ from relative extrema?
5. Can you explain why the endpoints of the interval are always checked?

Tip:

Always check the function's value at both the critical points and the endpoints when finding absolute extrema on a closed interval.