Question 18: Find all real values ​​of parameter m so that the function (y=frac{1}{3} x^{3}-mx^{2}+(m+1) x-1 ) peaks at x=-2?

\(y^{\prime}=x^{2}-2 m x+m+1, y^{\prime \prime}=2 x-2 m\)

The function reaches its maximum at x=-2 when: \(\left\{\begin{array}{l} y^{\prime}(-2)=0 \\ y^{\prime \prime}(-2)<0 \end{array} \Leftrightarrow\ left\{\begin{array}{l} 4+4 m+m+1=0 \\ 4-2 m<0 \end{array} \Leftrightarrow\left\{\begin{array}{l} m= -1 \\ m>2 \end{array}\right.\right.\right.\)(no solution)

So there is no value of m for the function to have an extreme at x=-2