To find the absolute maximum and minimum values of the function $f(x) = 2x^3 - x^2 - 4x + 6$ on the interval $[-1, 0]$, we'll follow these steps:

1. Find the derivative of $f(x)$ to determine the critical points where $f'(x) = 0$.
2. Evaluate the function at each critical point and at the endpoints $x = -1$ and $x = 0$.
3. Compare values to identify the absolute maximum and minimum values over the interval.

Step 1: Find the Derivative of $f(x)$

The function is: $f(x) = 2x^3 - x^2 - 4x + 6$ The derivative, $f'(x)$, is: $f'(x) = 6x^2 - 2x - 4$

Step 2: Solve $f'(x) = 0$ for Critical Points

Set $f'(x) = 0$: $6x^2 - 2x - 4 = 0$ To simplify, divide each term by 2: $3x^2 - x - 2 = 0$ This is a quadratic equation. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = -1$, and $c = -2$:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}$ $x = \frac{1 \pm \sqrt{1 + 24}}{6}$ $x = \frac{1 \pm \sqrt{25}}{6}$ $x = \frac{1 \pm 5}{6}$ Thus, we get: $x = 1 \quad \text{or} \quad x = -\frac{2}{3}$ Since we are restricted to the interval $[-1, 0]$, we only consider $x = -\frac{2}{3}$ as the critical point within this interval.

Step 3: Evaluate $f(x)$ at the Critical Point and Endpoints

Now, we evaluate $f(x)$ at $x = -1$, $x = -\frac{2}{3}$, and $x = 0$.

1. At $x = -1$: $f(-1) = 2(-1)^3 - (-1)^2 - 4(-1) + 6 = -2 - 1 + 4 + 6 = 7$

2. At $x = -\frac{2}{3}$: $f\left(-\frac{2}{3}\right) = 2\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) + 6$ Simplifying each term: $= 2 \left(-\frac{8}{27}\right) - \frac{4}{9} + \frac{8}{3} + 6$ $= -\frac{16}{27} - \frac{4}{9} + \frac{8}{3} + 6$ Converting to a common denominator (27): $= -\frac{16}{27} - \frac{12}{27} + \frac{72}{27} + \frac{162}{27} = \frac{206}{27} \approx 7.63$

3. At $x = 0$: $f(0) = 2(0)^3 - (0)^2 - 4(0) + 6 = 6$

Step 4: Determine the Maximum and Minimum Values

• The absolute maximum value is approximately $\frac{206}{27} \approx 7.63$, which occurs at $x = -\frac{2}{3}$.
• The absolute minimum value is $6$, which occurs at $x = 0$.

Final Answer

1. The absolute maximum value is $\frac{206}{27} \approx 7.63$ at $x = -\frac{2}{3}$.
2. The absolute minimum value is $6$ at $x = 0$.

Would you like further details on any of the steps?

1. How is the derivative of a function used to find critical points?
2. What is the significance of checking endpoints in an interval?
3. How do we solve quadratic equations in general?
4. Why do we use a common denominator to add fractions?
5. What are absolute maximum and minimum values in calculus?

Tip: Always verify that critical points lie within the interval before including them in your evaluations.