Question

Could somebody help my find the critical points for y=(x-3)(x+2)^(2/3). When I take the derivative of the function I get is x=-2,3. But when I put the original function in wolfram alpha there is minimum at x=0

Answer 1

your values of x = -2, 3 are the x-intercepts, and you don't need a derivative to find them

dy/dx = (x-3)(2/3)(x+2)^(-1/3) + (x+2)^(2/3)
= (1/3)(x+2)^(-1/3) [2(x-3) + 3(x+2)]
= (1/3)(5x)(x+2)^(-1/3)
= 0 for a max/min of y

----> 5x = 0 or (x+2)^(-1/3) = 0 ---> no way
x = 0, then y = -3(4)^(1/3) or appr -4.76

confirmed by Wolfram
http://www.wolframalpha.com/input/?i=plot+y+%3D+(x-3)(x%2B2)%5E(2%2F3)