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 :

Explanation

To solve the problem, we need to find the set \( S \) of all values of \( x \) for which the tangent to the curve \( y = f(x) = x^3 - x^2 - 2x \) at the point \( (x, f(x)) \) is parallel to the line segment joining the points \( (1, f(1)) \) and \( (-1, f(-1)) \).

### Step-by-step Solution:

1. **Calculate \( f(1) \) and \( f(-1) \)**:
\[ f(1) = 1^3 - 1^2 - 2 \cdot 1 = 1 - 1 - 2 = -2 \]
\[ f(-1) = (-1)^3 - (-1)^2 - 2 \cdot (-1) = -1 - 1 + 2 = 0 \]
Thus, the points are \( (1, -2) \) and \( (-1, 0) \).

2. **Find the slope of the line segment joining these points**:
The slope \( m \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-2)}{-1 - 1} = \frac{2}{-2} = -1 \]

3. **Find the derivative \( f'(x) \)**:
We differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - x^2 - 2x) = 3x^2 - 2x - 2 \]

4. **Set the derivative equal to the slope of the line segment**:
We want the tangent to the curve to have the same slope:
\[ f'(x) = -1 \]
Therefore, we set up the equation:
\[ 3x^2 - 2x - 2 = -1 \]
Simplifying gives:
\[ 3x^2 - 2x - 1 = 0 \]

5. **Solve the quadratic equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \]
This gives us two solutions:
\[ x = \frac{6}{6} = 1 \quad \text{and} \quad x = \frac{-2}{6} = -\frac{1}{3} \]

6. **Conclusion**:
The set \( S \) of all values of \( x \) for which the tangent to the curve is parallel to the line segment is:
\[ S = \{ 1, -\frac{1}{3} \} \]