To find the absolute maximum and minimum values of $f(x) = 4x^3 - 12x^2 - 36x + 5$ on the interval $[-2, 4]$, we follow these steps:

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

The derivative, $f'(x)$, will help identify critical points in the interval where the function may reach a maximum or minimum.

$f'(x) = \frac{d}{dx} (4x^3 - 12x^2 - 36x + 5) = 12x^2 - 24x - 36$

Step 2: Set the derivative to zero and solve for $x$

To find critical points, solve $f'(x) = 0$:

$12x^2 - 24x - 36 = 0$

Divide by 12:

$x^2 - 2x - 3 = 0$

Now, factor:

$(x - 3)(x + 1) = 0$

This gives two critical points:

$x = 3 \quad \text{and} \quad x = -1$

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

We need to evaluate $f(x)$ at $x = -2$, $x = 4$, $x = 3$, and $x = -1$ to find the absolute maximum and minimum.

1. At $x = -2$: $f(-2) = 4(-2)^3 - 12(-2)^2 - 36(-2) + 5 = -32 - 48 + 72 + 5 = -3$

2. At $x = 4$: $f(4) = 4(4)^3 - 12(4)^2 - 36(4) + 5 = 256 - 192 - 144 + 5 = -75$

3. At $x = 3$: $f(3) = 4(3)^3 - 12(3)^2 - 36(3) + 5 = 108 - 108 - 108 + 5 = -103$

4. At $x = -1$: $f(-1) = 4(-1)^3 - 12(-1)^2 - 36(-1) + 5 = -4 - 12 + 36 + 5 = 25$

Step 4: Determine the absolute maximum and minimum

Comparing the values:

• $f(-2) = -3$
• $f(4) = -75$
• $f(3) = -103$
• $f(-1) = 25$

The absolute minimum value is $-103$ (at $x = 3$), and the absolute maximum value is $25$ (at $x = -1$).

Answer

• Absolute Minimum Value: $-103$
• Absolute Maximum Value: $25$

Would you like more details on any step?

Here are five related questions to explore:

1. How would you determine absolute extrema for a function on an open interval?
2. What are possible methods to confirm whether a critical point is a local maximum or minimum?
3. How does changing the interval affect the absolute maximum and minimum values?
4. Can you apply this approach to find extrema for trigonometric functions?
5. How do we handle endpoints when finding extrema for a closed interval?

Tip: Always verify that your critical points lie within the given interval before evaluating the function at those points.