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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

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If only one line is tangent at two points, it means the gradient at -only - those two points are the same.

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.

You'll now have 2 co-ordinates; using Pythagoras' theorem you can find the distance between them.

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