Question

f(x)=6x^5-10x^3
Use derivatives to find the x-coordinates for all critical points of f and f'

Answer 1

f' = 30x^4 - 30x^2 = 0
--> 30x^2(x^2 -1) =0
--> x = 0, +-1

f'' = 120x^3 - 60x = 0
--> 60x(2x^2 -1) = 0
--> x = 0, +-sqrt2/2

Answer 2

Thank you so much!! :)