Question 40: Given the function \(y = mx⁴ + (m² – 6) x² + 4\) . How many integers m are there for the function to have three extremes, where there are exactly two minimums and one maximum?

TXĐ: \(D=\mathbb{R}\)

We have \(y’ = 4mx³ + 2 (m² – 6) x =0\Leftrightarrow 4x(mx^2+2(m^2-6))=0\)

\( \Leftrightarrow \left[\begin{array}{l}x=0\\m{x^2}+2({m^2}-6)=0\end{array}\right\)[\begin{array}{l}x=0\m{x^2}+2({m^2}-6)=0\end{array}\right\)

The given function has three extremes, where there are exactly two minimums and one maximum if and only if

\(\left\{ \begin{array}{l} 4m > 0\\ m\left( {{m^2} – 6} \right) < 0 \end{array} \right. \Leftrightarrow 0 < m < \sqrt 6 \)

Since m is an integer, there are 2 values of m that satisfy the problem condition

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