`f(x) = xsqrt(1 - x), [-1,1]` Find the local and absolute extreme values of the function on the given interval.

Textbook Question

Chapter 4, Review - Problem 2 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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], isĀ `2/(3sqrt3)` and it occurs at `x = 2/3` and the absolute minimum of the function is -sqrt2 and it occurs at x = -1.

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