If \(f(x) = 4x^3 - x^2 -2x+1\) and \(g(x) = \begin{cases}{\text{min} [f(t) : 0 \le t \le x]} && {;\>0\le\>x\le\>1} \\ {3-x} && {;\>1\lt\>x\>\le\>2}\end{cases} \)

Then find \(g\left(\frac{1}{4}\right) + g\left(\frac{3}{4}\right) + g\left(\frac{5}{4}\right)\).

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