Question

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

$g\left(x\right)=\{\begin{array}{cc}\text{Min.}\left\{f\right(t);0\le t\le x\};& 0\le x\le 1\\ 3-x& ;1\le x\le 2\end{array}$

then the value of $2\lambda$ if $\lambda =g\left(\frac{1}{4}\right)+g\left(\frac{3}{4}\right)+g\left(\frac{5}{4}\right)$ is

Answer 1

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



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

So $g\left(\frac{1}{4}\right)=f\left(\frac{1}{4}\right)=\frac{1}{2};g\left(\frac{3}{4}\right)=f\left(\frac{1}{2}\right)=\frac{1}{4};g\left(\frac{5}{4}\right)=3-\frac{5}{4}$

$\Rightarrow \lambda =\frac{1}{2}+\frac{1}{4}+3-\frac{5}{4}=\frac{5}{2}$

$\Rightarrow 2\lambda =5$