Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 2

You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0.
You need to evaluate the derivative using product rule, such that:
f'(x) = x'(sqrt(1-x)) + x*(sqrt(1-x))'
f'(x) = sqrt(1-x) - x/(2sqrt(1-x))
You need to solve for x the equation f'(x) =0:
sqrt(1-x) - x/(2sqrt(1-x)) = 0 => (2(1 - x) - x)/(2sqrt(1-x)) = 0 => 2(1 - x) - x = 0 => 2 - 2x - x = 0 => -3x = -2 => x = 2/3
You need to evaluate the function at critical points:
f(2/3) = (2/3)sqrt(1 - 2/3) => f(2/3) = 2/(3sqrt3)
You need to evaluate the function at the end points of interval:
f(-1) = (-1)sqrt(1+1) = > f(-1) = -sqrt2
f(1) = 1*sqrt(1-1) = 0
Hence, the absolute maximum of the function, on the interval [-1,1], is 2/(3sqrt3) and it occurs at x = 2/3 and the absolute minimum of the function is -sqrt2 and it occurs at x = -1.