Question

If $f\left(x\right)=4x^3-x^2-2x+1$ and $g\left(x\right)=\{\begin{array}{cc}min.\left\{f\right(t):0\le t\le x\};0\le x\le 1& \\ 3-x;1<x\le 2& \end{array}$ then the value of $\lambda$ lf $\frac{\lambda }{2}=g\left(\frac{1}{4}\right)+g\left(\frac{3}{4}\right)+g\left(\frac{5}{4}\right)$, is ___ .

Answer 1

$y=f\left(x\right)=4x^3-x^2-2x+1$

$g\left(x\right)=\{\begin{array}{ccc}f\left(x\right);0\le x\le \frac{1}{2}& & \\ f\left(\frac{1}{2}\right);\frac{1}{2}\le x\le 1& & \\ 3-x;1<x\le 2& & \end{array}Sog\left(\frac{1}{4}\right)=\frac{1}{2};g\left(\frac{3}{4}\right)=\frac{1}{4};g\left(\frac{5}{4}\right)=3-\frac{5}{4}.$