If for f(x) = 2x - x^2,Lagrange's Mean Value | vedclass

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(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 2x - x^

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$c = \frac{1}{2}$

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(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 2x - x^2$ on the interval $[0, 1]$,where $a = 0$ and $b = 1$.First,calculate $f(a)$ and $f(b)$:$f(0) = 2(0) - (0)^2 = 0$$f(1) = 2(1) - (1)^2 = 1$Next,calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x - x^2) = 2 - 2x$Now,substitute these values into the theorem formula:$f'(c) = \frac{f(1) - f(0)}{1 - 0}$$2 - 2c = \frac{1 - 0}{1}$$2 - 2c = 1$$2c = 1$$c = \frac{1}{2}$Thus,the value of $c$ is $\frac{1}{2}$.

If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

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