
Yefim S. answered 08/12/24
Math Tutor with Experience
f(x) = 5x + 9/x; f'(x) = 5 - 9/x2 = 0; x = ±3/√5
f''(x) = 18/x3; f''(3/√5) = 10√5/3 > 0. So, at x = 3/√5 f(x) has local minimum f(3/√5) = 6√5
f''(-3/√5) = -10√5/3 < 0. So, at x = -3/√5 f(x) has local minimum f(-3/√5) = -6√5.