Let f(x) = 8x^3 - 6x^2 - 2x + 1, then | vedclass

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(D) Let $F(x) = \int_0^x f(t) dt = \int_0^x (8t^3 - 6t^2 - 2t + 1) dt = 2x^4 - 2x^3 - x^2 + x.$We observe that $F(0) = 0$ and $F(1) = 2(1)^4 - 2(1)^3

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Both $(B)$ and $(C)$

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(D) Let $F(x) = \int_0^x f(t) dt = \int_0^x (8t^3 - 6t^2 - 2t + 1) dt = 2x^4 - 2x^3 - x^2 + x.$We observe that $F(0) = 0$ and $F(1) = 2(1)^4 - 2(1)^3 - (1)^2 + 1 = 2 - 2 - 1 + 1 = 0.$Since $F(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1).$By Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $F'(c) = f(c) = 0.$Thus,$f(x) = 0$ has at least one root in $(0, 1).$Also,since $f(0) = 1$ and $f(1) = 8 - 6 - 2 + 1 = 1,$ and $f(1/2) = 8(1/8) - 6(1/4) - 2(1/2) + 1 = 1 - 1.5 - 1 + 1 = -0.5,$ by the Intermediate Value Theorem,$f(x)$ has roots in $(0, 1/2)$ and $(1/2, 1).$Since $f(x)$ has at least two roots in $(0, 1),$ by Rolle's Theorem applied to $f(x),$ there exists some $c \in (0, 1)$ such that $f'(c) = 0.$Therefore,both $(B)$ and $(C)$ are correct.

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

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