To find the absolute minimum and maximum values of the function $f(x) = x^4 - 50x^2 + 5$ on the closed interval $-4 \leq x \leq 11$, we can follow these steps:

1. Compute the derivative $f'(x)$ and set it equal to zero to find critical points within the interval.
2. Evaluate $f(x)$ at each critical point and at the endpoints $x = -4$ and $x = 11$.
3. Compare these values to identify the minimum and maximum values.

Let's go through the solution:

Step 1: Find $f'(x)$ and Critical Points

The function is: $f(x) = x^4 - 50x^2 + 5$

The first derivative is: $f'(x) = 4x^3 - 100x$

To find the critical points, set $f'(x) = 0$: $4x^3 - 100x = 0$ $4x(x^2 - 25) = 0$ $x = 0, \quad x = 5, \quad x = -5$

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

Now we evaluate $f(x)$ at $x = -4$, $x = 11$, and the critical points $x = 0$, $x = 5$, and $x = -5$.

1. At $x = -4$: $f(-4) = (-4)^4 - 50(-4)^2 + 5 = 256 - 800 + 5 = -539$

2. At $x = 11$: $f(11) = (11)^4 - 50(11)^2 + 5 = 14641 - 6050 + 5 = 8596$

3. At $x = 0$: $f(0) = 0^4 - 50 \cdot 0^2 + 5 = 5$

4. At $x = 5$: $f(5) = (5)^4 - 50(5)^2 + 5 = 625 - 1250 + 5 = -620$

5. At $x = -5$: $f(-5) = (-5)^4 - 50(-5)^2 + 5 = 625 - 1250 + 5 = -620$

Step 3: Identify Absolute Minimum and Maximum

From the calculated values:

• The absolute minimum value is $-620$, occurring at $x = 5$ and $x = -5$.
• The absolute maximum value is $8596$, occurring at $x = 11$.

Final Answers

• Absolute minimum: $-620$
• Absolute maximum: $8596$

Would you like more details on the steps or have any questions?

Here are some related questions to deepen understanding:

1. How would the solution change if the interval was different?
2. What is the significance of critical points in finding absolute extrema?
3. Why is it necessary to evaluate endpoints in finding the absolute maximum and minimum on a closed interval?
4. Can there be more than one absolute maximum or minimum on a closed interval?
5. How would this problem change if we were only asked for local extrema?

Tip: Always remember to include endpoints when finding absolute extrema on a closed interval.