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

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(B) Given the function $f(x) = 2x^3 + ax^2 + bx$ on the interval $[-1, 1]$.According to Rolle's theorem,if $f(x)$ is continuous on $[-1, 1]$,different

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$-1$

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(B) Given the function $f(x) = 2x^3 + ax^2 + bx$ on the interval $[-1, 1]$.According to Rolle's theorem,if $f(x)$ is continuous on $[-1, 1]$,differentiable on $(-1, 1)$,and $f(-1) = f(1)$,then there exists at least one $c \in (-1, 1)$ such that $f'(c) = 0$.First,we use the condition $f(-1) = f(1)$:$f(1) = 2(1)^3 + a(1)^2 + b(1) = 2 + a + b$$f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) = -2 + a - b$Setting $f(1) = f(-1)$:$2 + a + b = -2 + a - b$$2b = -4 \implies b = -2$Next,we use the condition $f'(c) = 0$ at $c = \frac{1}{2}$:$f'(x) = 6x^2 + 2ax + b$$f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + 2a\left(\frac{1}{2}\right) + b = 0$$6\left(\frac{1}{4}\right) + a + b = 0$$\frac{3}{2} + a + b = 0$Substitute $b = -2$ into the equation:$\frac{3}{2} + a - 2 = 0$$a - \frac{1}{2} = 0 \implies a = \frac{1}{2}$Finally,calculate $2a + b$:$2a + b = 2\left(\frac{1}{2}\right) + (-2) = 1 - 2 = -1$

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

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