# Find the maximum value of $\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$

If $x\in\mathbb{R}$ find the maximum value of

$$\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$

I tried this:
Let $$y= \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$ For maxima $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2} < 0$. However, the equation $\frac{dy}{dx}=0$ (after simplifying and clearing the square roots) came out to be a nine degree equation which gave me a nightmare! Moreover, simplifying the derivative was also a tedious task. I found this question in my book in the chapter on theory of equation. I can't think of an algebraic solution. Please Help!
Thanks!

## 1 Answer

since $$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$

let $$P(x,x^2),A(3,2),B(0,1)$$ so $$|PA|-|PB|\le |AB|=\sqrt{10}$$ if and only is $A,P,B$ on a line.

• Very good solution! I like it a lot! – 5xum Jul 8 '14 at 16:59
• I don't see how this answers the question. What is the max? – Wintermute Jul 8 '14 at 16:59
• Yes very good.... but not very algebra-precalculus – Mathmo123 Jul 8 '14 at 17:00
• @china math Very Nice solution! I know to use circles for finding max/min of complex numbers and here you have used distance formula with triangle inequality. Can you give me some other examples where we can used this and some other shapes like hyperbola, parabola ellipse etc. to find extrema? Thanks! – Henry Jul 22 '14 at 16:53
• @Wintermute : The max is $\sqrt{10}$ and it occurs at one or both of the two points where the line $\overleftrightarrow{AB}$ intersects the parabola $y = x^2$. – steven gregory Aug 22 '16 at 21:57