The value of c in the Lagrange's mean value t | vedclass

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(D) According to Lagrange's Mean Value Theorem,there exists at least one $c \in (0, 1)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = x^

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$\frac{4-\sqrt{7}}{3}$

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(D) According to Lagrange's Mean Value Theorem,there exists at least one $c \in (0, 1)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = x^{3} - 4x^{2} + 8x + 11$ and interval $[0, 1]$,we have $a = 0$ and $b = 1$.Calculate $f(0) = 0^{3} - 4(0)^{2} + 8(0) + 11 = 11$.Calculate $f(1) = 1^{3} - 4(1)^{2} + 8(1) + 11 = 1 - 4 + 8 + 11 = 16$.The slope of the secant line is $\frac{f(1) - f(0)}{1 - 0} = \frac{16 - 11}{1} = 5$.Find the derivative $f'(x) = 3x^{2} - 8x + 8$.Set $f'(c) = 5$,so $3c^{2} - 8c + 8 = 5$,which simplifies to $3c^{2} - 8c + 3 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,we get $c = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}$.Since $c \in (0, 1)$,we choose $c = \frac{4 - \sqrt{7}}{3} \approx \frac{4 - 2.64}{3} \approx 0.45$,which lies in $(0, 1)$.

The value of $c$ in the Lagrange's mean value theorem for the function $f(x) = x^{3} - 4x^{2} + 8x + 11$ on the interval $x \in [0, 1]$ is:

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