Given,

$f_{\lambda }\left(x\right)=4\lambda x^3-36\lambda x^2+36x+48$

Differentiating w.r.t $x$ both side we get,

$f_{\lambda }^{\text{'}}\left(x\right)=12\lambda x^2-72\lambda x+36$

$f_{\lambda }{}^{\text{'}}\left(x\right)=12\left(\lambda x^2-6\lambda x+3\right)\ge 0$

$\therefore \lambda >0$ and $D\le 0$

$36{\lambda }^2-4\times \lambda \times 3\le 0$

$9{\lambda }^2-3\lambda \le 0$

$3\lambda \left(3\lambda -1\right)\le 0$

$\lambda \in \left[0,\frac{1}{3}\right]$

So, ${\lambda }_{\mathrm{largest} }=\frac{1}{3}$

$f\left(x\right)=\frac{4}{3}{x}^3-12x^2+36x+48$

So,$f\left(1\right)+f\left(-1\right)=72$