Key Concepts and Formulas

• Slope of a Tangent: The slope of the tangent line to a curve $y = f(x)$ at any point $(x, y)$ is given by its first derivative, $f'(x)$ or $\frac{dy}{dx}$.
• Slope of a Line Segment: The slope of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Parallel Lines: If two lines are parallel, their slopes are equal.
• Quadratic Formula: For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

1. Understand the Problem and Identify the Goal

The problem asks for the set $S$ of all $x$-values where the tangent to the curve $y = f(x) = x^3 - x^2 - 2x$ is parallel to the line segment connecting the points $(1, f(1))$ and $(-1, f(-1))$. This means the slope of the tangent at $x$ must equal the slope of the line segment. Our goal is to find these specific $x$-values.

2. Calculate the Coordinates of the Endpoints of the Line Segment

First, we need to find the exact coordinates of the two points that define the line segment. The given function is $f(x) = x^3 - x^2 - 2x$.

• For the point where $x=1$: $f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2$ So, the first point is $A(1, f(1)) = (1, -2)$.

• For the point where $x=-1$: $f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0$ So, the second point is $B(-1, f(-1)) = (-1, 0)$.

This step calculates the y-coordinates of the points on the curve at $x=1$ and $x=-1$, which are needed to find the slope of the secant line.

3. Determine the Slope of the Line Segment

Now, we calculate the slope of the line segment joining $A(1, -2)$ and $B(-1, 0)$. Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$: Let $(x_1, y_1) = (1, -2)$ and $(x_2, y_2) = (-1, 0)$. $m_{AB} = \frac{0 - (-2)}{-1 - 1} = \frac{2}{-2} = -1$ So, the slope of the line segment is $m_{AB} = -1$.

This step determines the slope of the secant line, which is the target slope for the tangent line.

4. Determine the Slope of the Tangent to the Curve

Next, we find the general expression for the slope of the tangent to the curve $y = f(x)$ at any point $(x, y)$. This is done by finding the first derivative of $f(x)$ with respect to $x$. Given $f(x) = x^3 - x^2 - 2x$. Differentiating term by term: $f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(x^2) - \frac{d}{dx}(2x)$ $f'(x) = 3x^2 - 2x - 2$ This expression, $3x^2 - 2x - 2$, represents the slope of the tangent to the curve at any point $x$.

This step calculates the derivative of the function, which represents the slope of the tangent line at any point on the curve.

5. Equate the Slopes and Formulate the Equation

The problem states that the tangent to the curve is parallel to the line segment. According to the property of parallel lines, their slopes must be equal. Therefore, we set the slope of the tangent ($f'(x)$) equal to the slope of the line segment ($m_{AB}$). $f'(x) = m_{AB}$ $3x^2 - 2x - 2 = -1$ Now, we rearrange this equation into a standard quadratic form $ax^2 + bx + c = 0$: $3x^2 - 2x - 2 + 1 = 0$ $3x^2 - 2x - 1 = 0$

This step sets the slope of the tangent equal to the slope of the secant, resulting in a quadratic equation that can be solved for $x$.

6. Solve the Quadratic Equation for $x$

We have a quadratic equation $3x^2 - 2x - 1 = 0$. We can solve this using factorization or the quadratic formula. Let's use factorization: $3x^2 - 3x + x - 1 = 0$ $3x(x - 1) + 1(x - 1) = 0$ $(3x + 1)(x - 1) = 0$

This gives us two possible values for $x$:

• $3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$
• $x - 1 = 0 \Rightarrow x = 1$

This step solves the quadratic equation to find the x-values where the tangent line has the desired slope.

7. State the Set $S$

The set $S$ consists of all values of $x$ for which the condition holds. Based on our calculations, these values are $1$ and $-\frac{1}{3}$. Therefore, $S = \left\{ -\frac{1}{3}, 1 \right\}$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating slopes and applying the quadratic formula. A small sign error can lead to incorrect solutions.
• Derivative Errors: Double-check your differentiation. Incorrect differentiation leads to the wrong slope for the tangent.
• Factorization Errors: Make sure your factorization is correct. Expand the factored form to confirm it matches the original quadratic equation.

Summary

We found the $x$-values for which the tangent to the curve $y = x^3 - x^2 - 2x$ is parallel to the line segment joining $(1, f(1))$ and $(-1, f(-1))$. We calculated the slope of the line segment to be -1, found the derivative of the function to be $3x^2 - 2x - 2$, equated them, and solved the resulting quadratic equation to find $x = -\frac{1}{3}$ and $x = 1$. Therefore, the set $S$ is $\left\{ -\frac{1}{3}, 1 \right\}$. This corresponds to option (B). However, the provided "Correct Answer" says (A) is correct. Since all the steps have been double-checked, I am still sticking to my solution. There might be a typo in the book.

Final Answer

The final answer is \boxed{\left{ -\frac{1}{3}, 1 \right}}, which corresponds to option (B).