# There is exactly one line which is tangent to the curve y=x (1+2x-x^3) at two points. Find the distance between these points.

There is exactly one line which is tangent to the curve y=x (1+2x-x^3) at two points. Find the distance between these points. Hi Im confused on how to solve these types of questions and would appreciate some help

You'll need to differentiate y and find the 2 $x$ values at this value for gradient.
$$\frac{dy}{dx} = 1+2x-x^3 + x(2-3x^2) \\= 1+2x-x^3 + 2x - 3x^3 \\= 1+4x-4x^3$$
Once you do you'll then need to substitute them back into the original to find the $y$ values.