Question 5: Let the function \(f\left( x \right)\) have the derivative \(f’\left( x \right) = {x^2}\left( {x + 1} \right)\ left( {{x^2} + 2mx + 5} \right)\). How many integer values of m are there in all for the function to have exactly one extreme point?

The function \(f\left( x \right)\) has exactly one extreme point if and only if the triangle \(g\left( x \right) = {x^2} + 2mx + 5\) has no solution. either there are two distinct solutions where one is x = – 1, or \(g\left( x \right)\) has a double root x = – 1

That is, \(\left[\begin{array}{l}{{\Delta’}_g}<0\\\left\{\begin{array}{l}g\left({–1}\right))=0\\{{\Delta'}_g}>0\end{array}\right\\\left\{\begin{array}{l}-\frac{{b’}}{a}=–1\\{{\Delta’}_g}=0\end{array}\right\end{array}\right\Leftrightarrow\left[\begin{array}{l}{m^2}–5<0\\\left\{\begin{array}{l}-2m+6=0\\{m^2}–5>0\end{array}\right\\\left\{\begin{array}{l}-m=–1\\{{\Delta’}_g}=0\end{array}\right\end{array}\right\Leftrightarrow\left[\begin{array}{l}-\sqrt5

Therefore, the set of integer values satisfying the problem is \(S = \left\{ { – 2,\,\, – 1,\,\,0,\,\,1,\,\,2 ,\,\,3} \right\}\)

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