To find the absolute maximum and minimum values of the function $f(x) = 3\sqrt{3}x^2 - 2x + 1$ on the interval $[-1, 2]$, we follow these steps:

Step 1: Find the derivative of the function

The derivative $f'(x)$ will help us identify any critical points (where $f'(x) = 0$ or is undefined) within the interval.

$f(x) = 3\sqrt{3}x^2 - 2x + 1$

Taking the derivative:

$f'(x) = 2 \cdot 3\sqrt{3}x - 2 = 6\sqrt{3}x - 2$

Step 2: Set the derivative equal to zero to find critical points

Now, solve $f'(x) = 0$:

$6\sqrt{3}x - 2 = 0$

Solving for $x$:

$6\sqrt{3}x = 2$

$x = \frac{2}{6\sqrt{3}} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9}$

This is the only critical point within the interval. Now, check if it lies within the interval $[-1, 2]$:

$\frac{\sqrt{3}}{9} \approx 0.192$

Since this value is between $-1$ and 2, it’s a valid critical point.

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

To find the absolute maximum and minimum values, evaluate $f(x)$ at the endpoints $x = -1$ and $x = 2$, as well as at the critical point $x = \frac{\sqrt{3}}{9}$.

1. At $x = -1$:

$f(-1) = 3\sqrt{3}(-1)^2 - 2(-1) + 1 = 3\sqrt{3}(1) + 2 + 1 = 3\sqrt{3} + 3$

Numerically:

$f(-1) = 3(1.732) + 3 = 5.196 + 3 = 8.196$

2. At $x = 2$:

$f(2) = 3\sqrt{3}(2)^2 - 2(2) + 1 = 3\sqrt{3}(4) - 4 + 1 = 12\sqrt{3} - 3$

Numerically:

$f(2) = 12(1.732) - 3 = 20.784 - 3 = 17.784$

3. At $x = \frac{\sqrt{3}}{9}$:

We approximate $x = 0.192$:

$f\left( \frac{\sqrt{3}}{9} \right) = 3\sqrt{3}\left( \frac{\sqrt{3}}{9} \right)^2 - 2\left( \frac{\sqrt{3}}{9} \right) + 1$

$f(0.192) \approx 1.011$

Step 4: Determine the absolute maximum and minimum

• $f(-1) = 8.196$
• $f(2) = 17.784$
• $f\left( \frac{\sqrt{3}}{9} \right) \approx 1.011$

The absolute maximum value is 17.784 at $x = 2$, and the absolute minimum value is 1.011 at $x = \frac{\sqrt{3}}{9}$.

Final answer:

• Absolute maximum: $17.784$ at $x = 2$
• Absolute minimum: $1.011$ at $x = \frac{\sqrt{3}}{9}$

Would you like further details or have any questions about this process?

Here are five related questions:

1. How do you find the derivative of more complex functions?
2. What is the significance of critical points in optimization problems?
3. How do we evaluate functions with irrational numbers?
4. What is the role of endpoints in finding absolute extrema?
5. How does the derivative test compare with other methods of finding extrema?

Tip: Always evaluate the function at both critical points and endpoints when searching for absolute extrema within a closed interval.