To find the absolute minimum and maximum values of the function $f(x) = x^8 + \frac{8}{x}$ on the interval $[1.5, 4]$, we need to follow these steps:

Step 1: Compute the Derivative

First, let's find the derivative $f'(x)$ to identify the critical points within the interval.

$f(x) = x^8 + \frac{8}{x}$

The derivative $f'(x)$ is:

$f'(x) = 8x^7 - \frac{8}{x^2}$

Step 2: Set the Derivative to Zero

We set $f'(x) = 0$ to find critical points.

$8x^7 - \frac{8}{x^2} = 0$

This can be rearranged to:

$8x^7 = \frac{8}{x^2}$

Dividing both sides by 8:

$x^7 = \frac{1}{x^2}$

Multiply both sides by $x^2$:

$x^9 = 1$

Taking the ninth root of both sides gives:

$x = 1$

However, $x = 1$ is not within the interval $[1.5, 4]$, so we discard this solution.

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

Since there are no critical points in the interval, we evaluate the function at the endpoints $x = 1.5$ and $x = 4$.

1. At $x = 1.5$: $f(1.5) = (1.5)^8 + \frac{8}{1.5}$ Calculating each part:

   • $(1.5)^8 = 25.62890625$
   • $\frac{8}{1.5} \approx 5.3333$

   So, $f(1.5) \approx 25.62890625 + 5.3333 = 30.9622$

2. At $x = 4$: $f(4) = (4)^8 + \frac{8}{4}$ Calculating each part:

   • $(4)^8 = 65536$
   • $\frac{8}{4} = 2$

   So, $f(4) = 65536 + 2 = 65538$

Conclusion

• The absolute minimum value of $f(x)$ on the interval $[1.5, 4]$ is $30.9622$ at $x = 1.5$.
• The absolute maximum value of $f(x)$ on the interval $[1.5, 4]$ is $65538$ at $x = 4$.

Would you like more details on this process or any specific step?

Follow-up Questions

1. What is the importance of checking endpoints when finding absolute extrema on a closed interval?
2. How would the process differ if the interval were open (e.g., $(1.5, 4)$)?
3. What would happen if a critical point did exist within the interval?
4. How do derivatives help identify local maxima and minima in general?
5. Can you explain why $x = 1$ was not considered a valid solution in this interval?

Tip: When solving for extrema on a closed interval, always check both critical points and endpoints.