Calculus:intervals of increase and decrease

F(x) = x square root (1-x)

Find the domain and the intervals of increase and decrease of f.

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f(x) = x√(1-x)            ← Notice that f(x) is real so long as √(1-x) is real.
                                     And, √(1-x) only if (1-x) is not negative.
                                     So, x is restricted by  1-x ≥ 0  so that  x ≤ 1

                                     Domain = {x│x ≤ 1}  OR  interval (-inf,1]            ← ANSWER

                  -x                               -x + 2[√(1-x)]²              2 - 3x
f'(x)  =  ———––  +  √(1-x)  =  ———––———–  =  ———–––
             2√(1-x)                                 2√(1-x)                 2√(1-x)


➊ f(x) is increasing where ever f'(x) is positive.
Notice that since square roots are never negative and division by 0 is prohibited,
the denominator of f'(x) ... i.e. 2√(1-x) ... is always positive.
It follows that f'(x) is positive whenever the numerator is positive.
So, we need  2-3x > 0  which requires  x < ⅔  (Note that this falls within the domain of f)

f(x) is positive for  x < ⅔ ... so that f(x) is increasing for  x < ⅔            ← ANSWER


➋ f(x) is decreasing where ever f'(x) is negative.
Again, the denominator of f'(x) ... i.e. 2√(1-x) ... is always positive.
It follows that f'(x) is negative whenever the numerator is negative.
So, we need  2-3x < 0  which requires  x > ⅔  
But, remember that the domain of f is restricted to a maximum of 1.

So, f(x) is negative for    ⅔ < x ≤ 1
so that f(x) is decreasing for  ⅔ < x ≤ 1            ← ANSWER



       ANSWER
Domain = (-inf,1]
f is increasing in (-inf,⅔)
f is decreasing in (⅔,1]


Have a good one!
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f(x) = x√(1 - x)

domain (- ∞, 1]

f ' (x) = -(x/2) (1 /√(1 - x)) + √(1 - x)

= 1/2√(1 - x)(-x + 2 - 2x) = ( 2 - 3x) /2√(1 - x)

equate f ' (x) to zero

2 - 3x = 0

=> x = 2/3

from (-∞ to 2/3) it increases and

from (2/3, 1) it decreases