Final answer:
The absolute maximum value of f(x) = x⁻² ln(x) on the interval [1/2, 5] is approximately 0.347 and occurs at x = 1/2. The absolute minimum value is approximately -0.366 and occurs at x = e.
Step-by-step explanation:
The function f(x) = x⁻² ln(x) is defined on the interval [1/2, 5]. To find the absolute maximum and minimum values of f on this interval, we can first find the critical points and then evaluate the function at these points and the endpoints.
To find the critical points, we need to find where the derivative of f(x) is equal to zero or undefined. Taking the derivative of f(x), we get f'(x) = -2x⁻³ ln(x) - x⁻².
For the interval [1/2, 5], the critical points are x = 1 and x = e (approximately 2.718), since f'(x) is undefined at x = 0 and negative for x < 1 and x > e, and positive for 1 < x < e.
Now we evaluate f(x) at these critical points and the endpoints:
- f(1/2) ≈ 0.347
- f(1) = ln(1) = 0
- f(e) ≈ -0.366
- f(5) ≈ -0.227
Therefore, the absolute maximum value of f on the interval [1/2, 5] is approximately 0.347 and occurs at x = 1/2, and the absolute minimum value is approximately -0.366 and occurs at x = e.