Let [] ^ * be the largest value offor which the functionf_ (x) = 4 x^3 - 36 x^2 + 36x + 48…

Question

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

Accepted Answer

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

$$\therefore$$ $$f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$$

For $${f_\lambda }(x)$$ increasing : $${(6\lambda )^2} - 12\lambda  \le 0$$

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

$$\therefore$$ $$\lambda^ * = {1 \over 3}$$

Now, $$f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$$

$$\therefore$$ $$f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$$

Solutions · 1