To find the absolute minimum and maximum values of the function $f(x) = x^4 - 18x^2 + 11$ over the interval $-2 \leq x \leq 7$, we need to follow these steps:

1. Compute the derivative $f'(x)$ and find the critical points by setting $f'(x) = 0$.
2. Evaluate $f(x)$ at the endpoints $x = -2$ and $x = 7$ and at any critical points within the interval $-2 \leq x \leq 7$.
3. Compare these values to determine the absolute minimum and maximum.

Let's proceed with these calculations.

Step 1: Compute $f'(x)$

Given $f(x) = x^4 - 18x^2 + 11$, the derivative $f'(x)$ is: $f'(x) = 4x^3 - 36x$

To find critical points, set $f'(x) = 0$: $4x^3 - 36x = 0$ Factor out $4x$: $4x(x^2 - 9) = 0$ $4x(x - 3)(x + 3) = 0$ This gives the solutions $x = 0$, $x = 3$, and $x = -3$.

Step 2: Evaluate $f(x)$ at Critical Points and Endpoints

Now we evaluate $f(x)$ at $x = -2$, $x = 7$, $x = 0$, $x = 3$, and $x = -3$, ensuring only points within $-2 \leq x \leq 7$ are considered.

Calculate $f(x)$:

1. At $x = -2$: $f(-2) = (-2)^4 - 18(-2)^2 + 11 = 16 - 72 + 11 = -45$

2. At $x = 7$: $f(7) = 7^4 - 18 \cdot 7^2 + 11 = 2401 - 882 + 11 = 1530$

3. At $x = 0$: $f(0) = 0^4 - 18 \cdot 0^2 + 11 = 11$

4. At $x = 3$: $f(3) = 3^4 - 18 \cdot 3^2 + 11 = 81 - 162 + 11 = -70$

5. At $x = -3$: $f(-3) = (-3)^4 - 18 \cdot (-3)^2 + 11 = 81 - 162 + 11 = -70$

Step 3: Identify Absolute Minimum and Maximum

From these values, we can see:

• The absolute minimum value is $-70$, occurring at $x = 3$ and $x = -3$.
• The absolute maximum value is $1530$, occurring at $x = 7$.

Final Answer

• Absolute minimum value: $-70$
• Absolute maximum value: $1530$

Would you like more details on any step, or have any questions?

Follow-Up Questions

1. What is the purpose of finding critical points in this context?
2. Why do we also evaluate the function at the endpoints of the interval?
3. How would we handle this problem if the interval were open (not including endpoints)?
4. What is the significance of the second derivative in determining the nature of critical points?
5. How would we approach the problem if there were no closed interval?

Tip

When solving for absolute extrema, always check both critical points and endpoints within the specified interval.