Min distance $f(x) = x^2+1$ and $g(x) = -x^2+12x-35$

[stuart clark]

find minimum distance between two parabolas f(x) = x²+1 and g(x) = -x²+12x-35

[senior]

Consider a point on x-y plane. How to find the smallest distance to a parabola?(Just draw)
Answer: When you find the closest point, it's obvious that the tangent line is perpendicular to the distance line you drew.

Similarly, while finding the closest point between two non-intersecting parabolas, we consider points of equal slope, more formally derivative. Let these points be (x1,y1) and (x2,y2). Then, the derivatives
f'(x1) = g'(x2) results in x1 = 6 - x2
The distance square is D = (y2-y1)² + (x2-x1)². We need this minimum, so take derivative and equate it to zero. We can substituting x1 = 6-x2 and find an equation only in terms of x1 or x2. I chose x1. So, D' turns out to be
D' = 2x1³ + x1 - 3 = 0, if you are careful enough you can see that x1 = 1 is a root of this equation.
So, x1 = 1, y1 = 2 and x2 = 5 and y2 = 0. The distance between these points is 2sqrt(5) (sqrt means square root)

[stuart clark]

Thanks Senior