The value of c in (0, 2) satisfying the Mean | vedclass

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(B) Given function $f(x) = x(x - 1)^2$ on the interval $[0, 2]$.According to the Mean Value Theorem,there exists at least one $c \in (0, 2)$ such that

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$4/3$

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(B) Given function $f(x) = x(x - 1)^2$ on the interval $[0, 2]$.According to the Mean Value Theorem,there exists at least one $c \in (0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = 0$ and $b = 2$.$f(0) = 0(0 - 1)^2 = 0$.$f(2) = 2(2 - 1)^2 = 2(1)^2 = 2$.So,$f'(c) = \frac{2 - 0}{2 - 0} = 1$.Now,find the derivative $f'(x)$:$f(x) = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$.$f'(x) = 3x^2 - 4x + 1$.Setting $f'(c) = 1$:$3c^2 - 4c + 1 = 1$.$3c^2 - 4c = 0$.$c(3c - 4) = 0$.This gives $c = 0$ or $c = 4/3$.Since $c$ must be in the open interval $(0, 2)$,we choose $c = 4/3$.

The value of $c$ in $(0, 2)$ satisfying the Mean Value Theorem for the function $f(x) = x(x - 1)^2, x \in [0, 2]$ is equal to

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