Parallel to x-axis

Given, $y={ x }^{ 3 }+{ x }^{ 2 }-x$ For tangents to be horizontal, slope of tangents = $0$ $\therefore \frac { dy }{ dx } =3{ x }^{ 2 }+2x-1=0$ Solving gives $x=-1$ and $x=\frac { 1 }{ 3 }$

Corresponding $y$ values are $y=1$ and $y=\frac { -5 }{ 27 }$

So it is clear that distance between two tangents is $[1-\frac { -5 }{ 27 } ]=\frac { 32 }{ 27 }$

FIRST UP,DIFFERENTIATE Y=F(X) W.R.T X NOW WE GET DY/DX = 3X^2 +2X-1 GIVEN SLOPE=0 THEREFORE, 3X^2+2X-1=0 WE GET X=1/3 , -1 NOW PUTTING X IN THE EQUATION WE GET Y= -5/27 , 1 HORIZONTAL LINE IS OF THE FORM Y=K FROM CALCULATED VALUES WE FIND DISTANCE = 1-(-5/27) = 32/27

This problem does not require Calculus to solve:

Let $f(x)=x^3+x^2-x$ . Let $y=d$ be the horizontal line tangent to the curve. Then, $f(x)=d$ so $x^3+x^2-x-d=0$ . Since, when graphed, $f(x)$ will be tangent to $y=d$ at $2$ distinct values of $d$ , there are $2$ distinct solutions for $f(x)=d$ . Thus, $x^3+x^2-x-d=0$ has $2$ distinct solutions.

Let $(x+b)(x+a)^2 = x^3+(2a+b)x^2+(a^2+2ab)x+(a^2b)$ represent $f(x)-d=0$ . Then, $(2a+b)=1$ and $(a^2+2ab)=-1$ . By substitution, simplification, and the Quadratic Formula, $(a,b)=(1,-1)$ and $(a,b)=(-0.333,1.6667)$ are obtained. Since $d=a^2b=-5/27$ and $d=1$ are the vertical positions of the horizontal lines. Thus, $1-(-5/7)=32/27$ which is the final answer.