\(f\left(x\right)=m\sqrt{x+1}-x+m\Rightarrow f'\left(x\right)=\frac{m}{2\sqrt{x+1}}-1=\frac{m-2\sqrt{x+1}}{2\sqrt{x+1}}\)
- Với \(m\ge6\Rightarrow f'\left(x\right)\ge0\) \(\forall x\in\left[3;8\right]\Rightarrow f\left(x\right)\) đồng biến
\(\Rightarrow f\left(x\right)_{max}=f\left(8\right)=4m-8=3\Rightarrow m=\frac{11}{4}\left(l\right)\)
- Với \(m\le6\Rightarrow f'\left(x\right)\le0\) \(\forall x\in\left[3;8\right]\Rightarrow f\left(x\right)\) nghịch biến
\(\Rightarrow f\left(x\right)_{max}=f\left(3\right)=3m-3=3\Rightarrow m=2\)
- Với \(6< m< 8\Rightarrow f'\left(x\right)=0\Leftrightarrow x=\frac{m^2-4}{4}\)
\(\Rightarrow f\left(x\right)_{max}=f\left(\frac{m^2-4}{4}\right)=m\left(1+\frac{m}{2}\right)-\frac{m^2-4}{4}=3\)
\(\Leftrightarrow m^2+4m-8=0\Rightarrow\left[{}\begin{matrix}m=-2-2\sqrt{3}\left(l\right)\\m=-2+2\sqrt{3}\left(l\right)\end{matrix}\right.\)
