Find the x coordinates of all relative extreme points of

Question

f(x)=1/4x^4 + 2/3x^3- 3/2x^2 + 4

f'(x)=x^3 + 2x^2 - 3x = 0

(x^2 + 3)(x-1) = 0

x=-3 x=1 x=0 Is this correct?

Answer 1

I get when factoring

x(x^2+2x-3)
x(x+3)(x-1)=0
x=0;-3, 1 same end as you, but your work is wrong.

Answer 2

thank you