Question 18: Find all values of parameter m so that the graph of the function y = x3-3( m+1) x2+ 12mx-3m+ 4 (C) has two extreme points A and B such that these two points together with the point C(-1; -9/2) form a triangle with the origin as the center of gravity.

We have the derivative y’ = 3x^{2}– 6(m + 1) x + 12m.

A function has two extremes if and only if y’ = 0 has two distinct solutions

Hay (m-1) ^{2 }> 0 deduce m≠1

^{Then the two extreme points are A( 2; 9m) : B( 2m; – 4m}3^{+ 12m}2

-3m+ 4).

\end{array}} \right. \Leftrightarrow m = – \frac{1}{2}\left( n \right)\)

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