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Find maximum of $|\sqrt{x^4-7x^2-4x+20}-\sqrt{x^4+9x^2+16}|$ without using any calculus

I believe I have made most of the progress but am stuck on the last step

I rewrote the expression as- $$\left|\sqrt{(x^2-4)^2+(x-2)^2}-\sqrt{(x^2+4)^2+(x-0)^2}\right|$$

We notice that this is written in the format of the distance formula. In particular, this is the expression for the difference of distances from an arbitrary point $(x^2,x)$ and $(4,2)$,$(-4,0)$ respectively.

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In this figure we have to find maximum of $|PA-PB|$

I could not think of how this could be maximised. Usually in questions like these $P$ somehow is collinear or lies on the reflection of some given point but nothing of that sort can happen here. I tried a few special points as well like $(4,-2),(0,0)$ etc.

Can someone help me with this?

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  • $\begingroup$ @TonyK yes, thanks for pointing it out $\endgroup$
    – Amadeus
    Sep 23, 2020 at 17:41

2 Answers 2

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The difference between the distances to each of those two points is maximised over the whole plane by any point on the straight line $AB$ that does not lie strictly between $A$ and $B$; and this maximum value is simply $|AB|=2\sqrt{17}$.

And your curve touches this line at $A$ when $x=2$; so the maximum value of the expression is $2\sqrt{17}$, achieved when $x=2$.

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  • $\begingroup$ I made a mistake while looking at the distance, I thought it was $7.2…$ instead of $8.2…>8$ I've edited my question so that someone in the future looking at this does not get confused; Thanks for your help $\endgroup$
    – Amadeus
    Sep 23, 2020 at 17:32
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By the triangle inequality, for every point $P$, $$||PA|-|PB||\leq |AB|$$ so the maximum is attained when $P=A$.

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