Let S be the set of all values of x for which | vedclass

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(D) First,calculate the coordinates of the points $(1, f(1))$ and $(-1, f(-1))$.$f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2$.$f(-1) = (-1)^3 - (-1)^

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$\left\{ -\frac{1}{3}, 1 \right\}$

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(D) First,calculate the coordinates of the points $(1, f(1))$ and $(-1, f(-1))$.$f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2$.$f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0$.The slope $m$ of the line segment joining $(1, -2)$ and $(-1, 0)$ is given by:$m = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{-2 - 0}{2} = -1$.The slope of the tangent to the curve $y = f(x)$ at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.$\frac{dy}{dx} = \frac{d}{dx}(x^3 - x^2 - 2x) = 3x^2 - 2x - 2$.Since the tangent is parallel to the line segment,their slopes must be equal:$3x^2 - 2x - 2 = -1$.$3x^2 - 2x - 1 = 0$.Factoring the quadratic equation:$3x^2 - 3x + x - 1 = 0$.$3x(x - 1) + 1(x - 1) = 0$.$(3x + 1)(x - 1) = 0$.Thus,$x = 1$ or $x = -\frac{1}{3}$.Therefore,$S = \left\{ -\frac{1}{3}, 1 \right\}$.

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f(x) = x^3 - x^2 - 2x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f(1))$ and $(-1, f(-1))$. Then $S$ is equal to:

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