Let λ ^ * be the largest value of λ for which the function f _ λ...

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,

${f}_{\lambda}^{'}\left(x\right) = 12\lambda{x}^{2} - 72\lambdax + 36$

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

$∴ \lambda > 0$ and $D \le 0$

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

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

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

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

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

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

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