If  $f\left(x\right)=x^3+4x^2+\lambda x+1$ is a monotonically decreasing function of $x$  in the largest possible interval
$(–2, –2/3)$. Then

Here,  $f^{\mathrm{\prime }}\left(x\right)\le 0$ $\Rightarrow$$3x^2+8x+\lambda \le 0\mathrm{\forall }x\in (-2,-\frac{2}{3})$

then situations for f'(x) are as follows;

Given that $f\left(x\right)$ decreases in the largest possible interval  $(-2,\frac{-2}{3})\text{.}$  then $f\text{'}\left(x\right)=0$ must have roots  –2 and  $–2/3$.Thus, product of roots 

 $(-2)(-\frac{2}{3})=\frac{\lambda }{3}$  $\Rightarrow$   $\lambda =4$