If f(x) = 4x^3 - x^2 - 2x + 1 and g(x) = begi | vedclass

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(D) Given $f(x) = 4x^3 - x^2 - 2x + 1$. First,we find the derivative: $f'(x) = 12x^2 - 2x - 2 = 2(6x^2 - x - 1) = 2(3x + 1)(2x - 1)$. For $x \in [0, 1

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$\frac{5}{2}$

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(D) Given $f(x) = 4x^3 - x^2 - 2x + 1$. First,we find the derivative: $f'(x) = 12x^2 - 2x - 2 = 2(6x^2 - x - 1) = 2(3x + 1)(2x - 1)$. For $x \in [0, 1]$,$f'(x) = 0$ at $x = \frac{1}{2}$. $f(x)$ decreases on $[0, \frac{1}{2}]$ and increases on $[\frac{1}{2}, 1]$. For $g(x) = \min \{f(t) : 0 \le t \le x\}$ when $0 \le x \le 1$: If $0 \le x If $\frac{1}{2} \le x \le 1$,$g(x) = f(\frac{1}{2}) = 4(\frac{1}{8}) - \frac{1}{4} - 2(\frac{1}{2}) + 1 = \frac{1}{2} - \frac{1}{4} - 1 + 1 = \frac{1}{4}$. Now calculate the values: $g(\frac{1}{4}) = f(\frac{1}{4}) = 4(\frac{1}{64}) - \frac{1}{16} - 2(\frac{1}{4}) + 1 = \frac{1}{16} - \frac{1}{16} - \frac{1}{2} + 1 = \frac{1}{2}$. $g(\frac{3}{4}) = f(\frac{1}{2}) = \frac{1}{4}$ (since $\frac{3}{4} \ge \frac{1}{2}$). $g(\frac{5}{4}) = 3 - \frac{5}{4} = \frac{12-5}{4} = \frac{7}{4}$. Sum $= g(\frac{1}{4}) + g(\frac{3}{4}) + g(\frac{5}{4}) = \frac{1}{2} + \frac{1}{4} + \frac{7}{4} = \frac{2+1+7}{4} = \frac{10}{4} = \frac{5}{2}$.

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

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