Official solution
Given f(x) = x3 -- x2 -- 2x
\[ \therefore \] f(1) = 1 -- 1 -- 2 = - 2
and f(-1) = -1 -- 1 + 2 = 0
So point A(1, f(1)) = (1, -2)
and point B(--1, f(--1)) = (-1, 0)
Slope of tangent at point (x, y) to the curve
y = f(x) = x3 -- x2 -- 2x is
\[{{dy} \over {dx}}\] = 3x2 - 2x - 2
Slope of line segment joining the
points (1, -2) and (--1, 0) is
= \[{{ - 2 - 0} \over {1 - \left( { - 1} \right)}}\]
As tangent to the curve and line segment both are parallel then slope of them are same.
\[ \therefore \] 3x2 - 2x - 2 = \[{{ - 2 - 0} \over {1 - \left( { - 1} \right)}}\]
\[ \Rightarrow \] 3x2 - 2x - 1 = 0
\[ \Rightarrow \] x = \[{{2 \pm \sqrt {4 - 4.3\left( { - 1} \right)} } \over {2.3}}\]
= \[{{2 \pm 4} \over 6}\]
\[ \therefore \] x = 1, \[ - {1 \over 3}\]
So, S = \[\left\{ { - {1 \over 3}, 1} \right\}\]