Find the absolute extrema of the function on the closed interval. g(x) = lnx/5x , [1, 4]

Answer 1

g(x) = lnx / (5x)

g'(x) = [5x(1/x) - 5lnx] / (25x²) = [5 - 5lnx] / (25x²)

g'(x) = 0 when 5 - 5lnx = 0 So, lnx = 1. Therefore, x = e is the only critical point in [1,4].

Absolute extrema of a continuous function on a closed interval can occur only at an endpoint or critical point.

g(1) = 0

g(e) = 1/(5e) ≈ 0.074 (absolute max)

g(4) = ln4 / 20 ≈ 0.069 (absolute min)