Question

Consider the function f(x) = (3/3)x - (x-1)^(1/5) on the interval [1, [infinity]). Find the relative and absolute extrema. Use the first derivative test if needed. No graph is necessary.

Answer 1

The function has a minimum relative extremum at x ≈ 1.364 and the absolute extrema occur at the endpoints of the interval.

Step-by-step explanation:

To find the relative and absolute extrema of the function f(x) = (3/5)x - (x-1)^1/5 on the interval [1, [infinity]), we need to find the critical points and endpoints of the function and evaluate them.

First, let's find the critical points by taking the derivative of the function:

f'(x) = 3/5 - (1/5)(x-1)^-4/5

Setting f'(x) = 0 and solving for x, we get:

3/5 - (1/5)(x-1)^-4/5 = 0

Simplifying, we find:

(x-1)^-4/5 = 3/5

(x-1) = [(3/5)^5/4]

x = [(3/5)^5/4] + 1

Using a calculator, we find x ≈ 1.364.

Now, let's evaluate the function at the critical points and endpoints:

f(1) = (3/5)(1) - (1-1)^1/5 = 3/5 - 0 = 3/5

f(∞) = (3/5)(∞) - (∞-1)^1/5 = ∞ - ∞ = undefined

The function has a minimum relative extremum at x ≈ 1.364, and the absolute extrema occur at the endpoints of the interval: f(1) = 3/5 and f(∞) = undefined.