Question

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

Answer 1

$y=f\left(x\right)=x^3-x^2-2x\Rightarrow f^{\prime }\left(x\right)=3x^2-2x-2\cdots \left(i\right)$
Also $f\left(1\right)=1-1-2=-2$ and
$f(-1)=-1-1+2=0\Rightarrow m=\frac{f\left(1\right)-f(-1)}{1+1}=\frac{-2-0}{2}\Rightarrow m=-1\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$3x^2-2x-2=-1\Rightarrow 3x^2-2x-1=0\Rightarrow (3x+1)(x-1)=0\Rightarrow x=-\frac{1}{3},1$