Question 14: Find the values of the parameter m so that the graph of the function: \(y = {x^4} – 2m{x^2} + 2m + {m^4}\) has three extreme points of three vertex of an equilateral triangle

\(\begin{array}{l}

y’ = 4{x^3} – 4mx\\

y’ = 0 \Leftrightarrow 4x\left( {{x^2} – m} \right) = 0

\end{array}\)

Function with 3 extremes m > 0

Then the three extreme points of the graph of the function are

\(A\left( {0;{m^4} + 2m} \right),B\left( { – \sqrt m ;{m^4} – {m^2} + 2m} \right),C \left( {\sqrt m ;{m^4} – {m^2} + 2m} \right)\)

Due to symmetry, we have ABCC isosceles at the vertex A

So ∆ABC both need only AB = BC

\( \Leftrightarrow m + {m^4} = 4m \Leftrightarrow \left[ \begin{array}{l}

m = 0\\

m = \sqrt[3]{3}

\end{array} \right.\)

Combining the conditions we have \(m = \sqrt[3]{3}\) (satisfied)

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