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

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

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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(52780) (Show Source): You can put this solution on YOUR website!
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How this problem is worded, printed and posted, it does not seem to be a correct formulation.

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