Ta có: \(y'=\left(m+1\right)x^2+2\left(2m-1\right)x-\left(3m+2\right)\)
Để hàm số nghịch biến trên khoảng có độ dài bằng 4 thì y'=0 phải có hai nghiệm phân biệt, (m+1)>0 và \(\left|x_1-x_2\right|=4\)
Ta có: m+1>0 \(\Rightarrow m>-1\) (1)
TH1: m=-1
\(\Rightarrow y'=-6x+1=0\Rightarrow x=\dfrac{1}{6}\) (loại)
TH2: \(m\ne-1\)
\(\Rightarrow\Delta=4\left(2m-1\right)^2+4\left(3m+2\right)\left(m+1\right)\)
\(\Leftrightarrow\Delta=4\left(4m^2-4m+1\right)+4\left(3m^2+3m+2m+2\right)\)
\(\Leftrightarrow\Delta=16m^2-16m+4+12m^2+20m+8\)
\(\Leftrightarrow\Delta=28m^2+4m+12>0\forall m\in R\)
Theo định lí Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2\left(2m-1\right)}{\left(m+1\right)}\\x_1x_2=\dfrac{-\left(3m+2\right)}{\left(m+1\right)}\end{matrix}\right.\)
Ta có: \(\left|x_1-x_2\right|=4\Rightarrow\left(x_1-x_2\right)^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)
\(\Leftrightarrow\left(\dfrac{-2\left(2m-1\right)}{m+1}\right)^2+\dfrac{4\left(3m+2\right)}{m+1}=16\)
\(\Leftrightarrow\dfrac{4\left(2m-1\right)^2}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)
\(\Leftrightarrow\dfrac{4\left(4m^2-4m+1\right)}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)
\(\Leftrightarrow\dfrac{16m^2-16m+4}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)
\(\Leftrightarrow16m^2-16m+4+\left(m+1\right)\left(12m+8\right)=16\left(m+1\right)^2\)
\(\Leftrightarrow16m^2-16m+4+12m^2+8m+12m+8=16m^2+32m+16\)
\(\Leftrightarrow12m^2-28m-4=0\)
\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{7+\sqrt{61}}{6}\\m=\dfrac{7-\sqrt{61}}{6}\end{matrix}\right.\)(TM)
