How do use the first derivative test to determine the local extrema #y=x(sqrt(8-x^2))#?

1 Answer
Aug 13, 2015

Answer:

# x = +- 2, y = +-4 #

Explanation:

A stationary point can be obtained from # f'(x) = 0 #

# y = xsqrt(8-x^2) #
# y' = sqrt(8-x^2) - x^2/sqrt(8-x^2) = (-2(x^2-4))/sqrt(8-x^2) #

# (-2(x^2-4))/sqrt(8-x^2) = 0 #
# x^2 = 4 #
# x = +- 2, y = +-4 #

You can then proceed to find whether these points are local maxima or minima by taking the second derivative and substituting the values of the stationary points, #alpha#:
#f''(alpha) > 0 rArr# minima
#f''(alpha) < 0 rArr# maxima

# y'' = -(2x(12-x^2))/(8-x^2)^(3/2) #
# y''_(x=2) = -4 < 0 rArr # maxima
# y''_(x=-2) = 4 > 0 rArr # minima

graph{xsqrt(8-x^2) [-5, 5, -6, 6]}