As a real function:
#y = xsqrt(1-x^2)#
is defined when #1-x^2 >= 0#, that is for #x in [-1,1]#.
So the function is continuous on a compact domain and Weierstrass' theorem ensures it has an absolute maximum and an absolute minimum.
Evaluate first the value of #y# and the limits of the domain:
#y(-1) = y(1) = 0#
Evaluate now the first derivative:
#dy/dx = x (d/dx sqrt(1-x^2)) + (d/dx x) sqrt(1-x^2)#
#dy/dx = x (-x/sqrt(1-x^2)) + sqrt(1-x^2)#
#dy/dx = sqrt(1-x^2) -x^2/sqrt(1-x^2)#
#dy/dx = (1-x^2 -x^2)/sqrt(1-x^2)#
#dy/dx = (1-2x^2 )/sqrt(1-x^2)#
and find the critical points of #y(x)# inside its domain solving the equation:
#dy/dx =0#
#(1-2x^2 )/sqrt(1-x^2) = 0#
#1-2x^2 = 0#
#x=+-1/sqrt2#
Looking at the expression of #y'(x)# we can see that the denominator is always positive, while the numerator is a second degree polynomial with leading negative coefficient. #y'(x)# is therefore negative in the interval outside the roots and positive in the interval between the roots.
We can then conclude that:
#y(x)# is decreasing in #(-1,-1/sqrt2)#
#y(x)# is increasing in #(-1/sqrt2,1/sqrt2)#
#y(x)# is decreasing in #(1/sqrt2,1)#
so that #x=-1/sqrt2# is a local minimum and #x=1/sqrt2# is a local maximum. Evaluating:
#y(-1/sqrt2) = (-1/sqrt2)sqrt(1-1/2) = -1/2 #
#y(1/sqrt2) = (1/sqrt2)sqrt(1-1/2) = 1/2 #
we can see that the minimum is lower and the maximum is higher than the values of #y(x)# at the boundaries, so these are also the absolute minimum and maximum.
