Thanks!

Distance between parabolas : (x^2+1)-(x-x^2)

This equals :

x^2+1-x+x^2

2x^2-x+1

Let’s find the derivate F'(x)

4x-1

And its 0 :

4x-1 = 0

4x=1

1/4=x

Therefore, the minimum vertical distance between the parabolas is when X=1/4

The distance is F'(1/4) = 0.875

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