Let ${\lambda }^{*}$ be the largest value of $\lambda$ for which the function $f_{\lambda }\left(x\right)=4\lambda x^3-36\lambda x^2+36x+48$ is increasing for all $x\in \mathrm{ℝ}$. Then $f_{{\lambda }^{*}}\left(1\right)+f_{{\lambda }^{*}}(-1)$ is equal to :

$\begin{array}{l}{\mathrm{f}}_{\lambda }\left(\mathrm{x}\right)=4\lambda {\mathrm{x}}^3-36\lambda {\mathrm{x}}^2+36\mathrm{x}+48\\ {{\mathrm{f}}_{\lambda }}^{\mathrm{\prime }}\left(\mathrm{x}\right)=12\lambda {\mathrm{x}}^2-72\lambda \mathrm{x}+36\\ {{\mathrm{f}}_{\lambda }}^{\mathrm{\prime }}\left(\mathrm{x}\right)=12\left(\lambda {\mathrm{x}}^2-6\lambda \mathrm{x}+3\right)\ge 0\\ \therefore \lambda >0\mathrm{\&}\mathrm{D}\le 0\\ 36{\lambda }^2-4\times \lambda \times 3\le 0\\ 9{\lambda }^2-3\lambda \le 0\\ 3\lambda (3\lambda -1)\le 0\\ \begin{array}{l}\lambda \in \left[0,\frac{1}{3}\right]\\ \therefore {\lambda }_{\text{largest }}=\frac{1}{3}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{4}{3}{\mathrm{x}}^3-12{\mathrm{x}}^2+36\mathrm{x}+48\\ \therefore \mathrm{f}*\left(1\right)+\mathrm{f}*(-1)=72\end{array}\end{array}$