To solve for 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: Find the derivative of $f(x)$

The function is: $f(x) = x^8 + \frac{8}{x}$ To find the critical points, we need to calculate $f'(x)$ and set it equal to zero.

1. Differentiate $x^8$: $\frac{d}{dx} (x^8) = 8x^7$

2. Differentiate $\frac{8}{x}$: $\frac{d}{dx} \left( \frac{8}{x} \right) = -\frac{8}{x^2}$

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

Step 2: Set $f'(x) = 0$ to find critical points

$8x^7 - \frac{8}{x^2} = 0$ Divide by 8: $x^7 - \frac{1}{x^2} = 0$ Multiply by $x^2$ to eliminate the fraction: $x^9 = 1$ Taking the ninth root of both sides, we find: $x = 1$ Since $x = 1$ is not in the interval $[1.5, 4]$, we disregard it.

Step 3: Evaluate $f(x)$ at the endpoints of the interval

We evaluate $f(x)$ at $x = 1.5$ and $x = 4$ since there are no critical points in the interval.

1. At $x = 1.5$: $f(1.5) = (1.5)^8 + \frac{8}{1.5} = 25.62890625 + 5.333333 = 30.96223925$

2. At $x = 4$: [ f(4) = 4^8 + \frac{8}{4} = 65536 + 2 =