Question

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

Answer 1

`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]}