Chọn A
Ta có\( y=g\left(x\right)=f\left(x^{3} +3x\right)+m\Rightarrow g'\left(x\right)=\left(3x^{2} +3\right)f'\left(x^{3} +3x\right)\)
Do đó\( g'\left(x\right)=0\Leftrightarrow f'\left(x^{3} +3x\right)=0\Leftrightarrow \left[\begin{array}{l} {x^{3} +3x=-3} \\ {x^{3} +3x=-\frac{4}{3} } \\ {x^{3} +3x=0} \\ {x^{3} +3x=2} \end{array}\right. \Leftrightarrow \left[\begin{array}{l} {x\approx -0,82} \\ {x\approx -0,42} \\ {x=0} \\ {x\approx 0,6} \end{array}\right. \)
Bảng biến thiên
![](https://ryl16zv916obj.vcdn.cloud/hoidap/?qa=blob&qa_blobid=1022763286773667671)
Từ BBT suy ra \({\mathop{\max g\left(x\right)}\limits_{\left[0;1\right]}} =3+m=4\Leftrightarrow m=1.\) Và \({\mathop{\min g\left(x\right)}\limits_{\left[-1;0\right]}} =-1+m=-2\Leftrightarrow m=-1\)
Vậy \(m_{1} +m_{2} =0.\)