How do you determine the absolute extreme values for the function #y=x(sqrt(1-x²))# on its domain?

Question

Ezra Adams · Accepted Answer

The absolute minimum is:

#y(-1/sqrt2) = -1/2#

and the absolute maximum is:

#y(1/sqrt2) = 1/2# 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.

Paisley Johnson · Answer

undefined To determine the absolute extreme values for the function $ y = x \sqrt{1 - x^2} $ on its domain, follow these steps:

1. Find the critical points by taking the derivative of the function and setting it equal to zero.
2. Determine the endpoints of the domain.
3. Evaluate the function at the critical points and endpoints.
4. The largest and smallest values obtained in step 3 are the absolute maximum and minimum values, respectively, of the function on its domain.