SOLUTION: g(x)=f(x^3-3x^2+1) so how many extreme points y = f (x). X=0 2 3
G’(x)= -0+0-0+
G(x)= positive infinity 2 18 2 positive infinity
Algebra ->
Equations
-> SOLUTION: g(x)=f(x^3-3x^2+1) so how many extreme points y = f (x). X=0 2 3
G’(x)= -0+0-0+
G(x)= positive infinity 2 18 2 positive infinity
Log On
Question 1165104: g(x)=f(x^3-3x^2+1) so how many extreme points y = f (x). X=0 2 3
G’(x)= -0+0-0+
G(x)= positive infinity 2 18 2 positive infinity Answer by ikleyn(52778) (Show Source):
