Find the minimum of $f(x)=\sqrt{(1-x^2)^2+(2-x)^2}+\sqrt{x^4-3x^2+4}$.
I haven't learned derivative so I tried to solve it with geometry.But in the end I failed.
Maybe it can be seemed as the sum of the lengths of two line segments?
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$$f(x)=\sqrt{(x-2)^2+(x^2-1)^2}+\sqrt{(x)^2+(x^2-2)^2}$$
So $f(x)$ is the sum of the distances from a point on the parabola $y=x^2$ to points $(2,1)$ and $(0,2)$.
$(0,2)$ is inside the parabola and $(2,1)$ is outside, so the point $(x,x^2)$ that makes $f(x)$ minimum would lie on the line segment that connects the two points.
Therefore, the minimum of $f(x)$ is $\sqrt{5}$.
Saturday
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Why is it important that one point is in-, the other outside? – mafu Mar 06 '22 at 12:03
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@mafu If they're both inside or both outside, the line that connects the two points would have to "bounce off" the parabola, so it wouldn't be a simple line segment. This requires different calculations. – Saturday Mar 06 '22 at 13:36