PLS HALP

Let P be denoted as  (x, 0)

So PA  =    sqrt [ ( -2 -x)^2  + (-1)^2 ]    =  sqrt [ 4 + 4x + x^2 + 1]  =  sqrt [ x^2 + 4x + 5]

And PB  =  sqrt [ (-4 -x)^2  + 2^2 ]  =  sqrt [ 16 + 8x + x^2 + 4 ]  = sqrt [ x^2 + 8x + 20 ]

And we wish to maximize D.....where D  =  

abs [sqrt(x^2 + 4x + 5)  -  sqrt ( x^2 + 8x + 20) ]    

Take the derivative of the function inside the absolute value bars and and set to 0

[x + 2] / sqrt (x^2 + 4x + 5)   -   [x + 4] /  sqrt (x^2 + 8x + 20)    = 0

[x + 2] / sqrt (x^2 + 4x + 5)   =   [x + 4] /  sqrt (x^2 + 8x + 20)    cross-multiply

[x + 2]sqrt(x^2 + 8x + 20)   =  [x + 4]sqrt(x^2 + 4x+ 5)        square both sides

[x^2 + 4x + 4] [x^2 + 8x + 20]   = [x^2 + 8x + 16] [x^2 + 4x + 5]   simplify

x^4 + 12 x^3 + 56 x^2 + 112 x + 80   = x^4 + 12 x^3 + 53 x^2 + 104 x + 80

56x^2 + 112x   =  53x^2  + 104x

3x^2 + 8x  =  0       factor

 x (3x + 8) = 0

There are two possible solutions......x  = 0    or x  = -8/3

And this graph confirms that for the original function,

abs [sqrt(x^2 + 4x + 5)  -  sqrt ( x^2 + 8x + 20) ], 

the absolute value of PA -PB  is greatest ( = √5 ) when  x =0  

https://www.desmos.com/calculator/vvmfqwf6wv

So.....(0,0)  is the point that maximizes the absolute value of PA - PB

{Thanks to heureka for pointing out my previous mistake...]

http://www.math.harvard.edu/~knill/teaching/math21a2011/21a_fall_2011.pdf