The constant c of Rolle's theorem for the fun | vedclass

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(D) For Rolle's theorem to be applicable,$f(x)$ must be continuous on $[1, 2]$,differentiable on $(1, 2)$,and $f(1) = f(2)$.Here,$f(1) = (1-1)^3(1-2)^

Vedclass · Accepted Answer

$\frac{11}{8}$

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(D) For Rolle's theorem to be applicable,$f(x)$ must be continuous on $[1, 2]$,differentiable on $(1, 2)$,and $f(1) = f(2)$.Here,$f(1) = (1-1)^3(1-2)^5 = 0$ and $f(2) = (2-1)^3(2-2)^5 = 0$. Since $f(1) = f(2) = 0$,Rolle's theorem applies.We need to find $c \in (1, 2)$ such that $f'(c) = 0$.Using the product rule: $f'(x) = 3(x-1)^2(x-2)^5 + 5(x-1)^3(x-2)^4$.Factor out common terms: $f'(x) = (x-1)^2(x-2)^4 [3(x-2) + 5(x-1)]$.$f'(x) = (x-1)^2(x-2)^4 [3x - 6 + 5x - 5] = (x-1)^2(x-2)^4 (8x - 11)$.Setting $f'(c) = 0$ for $c \in (1, 2)$:$(c-1)^2(c-2)^4 (8c - 11) = 0$.Since $c 
eq 1$ and $c 
eq 2$,we must have $8c - 11 = 0$,which gives $c = \frac{11}{8}$.

The constant $c$ of Rolle's theorem for the function $f(x)=(x-1)^3(x-2)^5$ in the interval $[1, 2]$ is:

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