Mark M. answered 05/29/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x) = x √(2-x) Domain: 2-x > 0. So, x < 2
f'(x) = √(2-x) - x(1/2)(2-x)-1/2 = √(2-x) - x / (2√2-x) = [2(2-x) - x] / (2√2-x) = (4-3x) / [2√(2-x)]
f'(x) = 0 when x = 4/3
When x < 4/3, f'(x) > 0. So, f(x) is increasing
When 4/3 < x < 2, f'(x) < 0. So, f is decreasing