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Question

Find the derivative at each critical point and determine local extreme values of y=x^2 * the square root of (1-x)

Answer 1

y'=[1/(2(1-x)^.5][3x^2+4x]
roots are
x=0 and x=-4/3 and x not equal to1

Take the second derivative an dsubstiture the roots. If y" <0 relative maximum, y">0 relative minumum

Answer 2

Y=x*x*sqrt(1-x), 1-x must be >= 0, thus x<=1
Y does not exist for x>1
Y’=2*x*sqrt(1-x)-x*x/2/sqrt(1-x) = x*(2-2.5x)/sqrt(1-x)
The Y’(x=1) does no exist.
Y’=0 at x1=0 & x2=0.8, that is OK with x<=1
Y’(0)=0, local min
Y’(0.8)=0.286 – local max
Y(x=1)=0 edge point

Answer 3

y = x^2 times (1-x)^.5

dy/dx=x^2 times -.5(1-x)^-.5 + 2x(1-x)^.5
=-x^2/(1-x)^.5 + 2x(1-x)^.5
-x/(1-x)^.5 + 2(1-x)^.5
Now set above = 0 and solve for x.
-x +2(1-x) =0
-3x=-2
x=2/3