If Rolle's theorem holds for the function f(x | vedclass

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(B) For Rolle's theorem to hold on $[-1, 1]$,we must have $f(-1) = f(1)$.$f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = -2 + b - c$$f(1) = 2(1)^3 + b(1)^2 + c(1

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(B) For Rolle's theorem to hold on $[-1, 1]$,we must have $f(-1) = f(1)$.$f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = -2 + b - c$$f(1) = 2(1)^3 + b(1)^2 + c(1) = 2 + b + c$Equating $f(-1) = f(1)$:$-2 + b - c = 2 + b + c$$-2c = 4 \implies c = -2$Since Rolle's theorem holds at $x = \frac{1}{2}$,we have $f'(\frac{1}{2}) = 0$.$f'(x) = 6x^2 + 2bx + c$$f'(\frac{1}{2}) = 6(\frac{1}{4}) + 2b(\frac{1}{2}) + c = 0$$\frac{3}{2} + b + c = 0$Substituting $c = -2$:$\frac{3}{2} + b - 2 = 0 \implies b = 2 - \frac{3}{2} = \frac{1}{2}$Now,calculate $(2b + c)$:$2b + c = 2(\frac{1}{2}) + (-2) = 1 - 2 = -1$.

If Rolle's theorem holds for the function $f(x) = 2x^3 + bx^2 + cx$ on the interval $x \in [-1, 1]$ at the point $x = \frac{1}{2}$,then $(2b + c)$ is equal to:

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