How do you express # y=|x^2-9|+|x^4-16|+|x^6-1|#, sans the symbol #|...|#?

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Answer 1 · 2016-12-14T17:45:31

#y = sqrt((x^2-9)^2)+sqrt((x^4-16)^2)+sqrt((x^6-1)^2)#

#abs(x) = sqrt(x^2)#

This is a one dimensional version of the distance formula.

So:

#y = sqrt((x^2-9)^2)+sqrt((x^4-16)^2)+sqrt((x^6-1)^2)#

graph{y = sqrt((x^2-9)^2)+sqrt((x^4-16)^2)+sqrt((x^6-1)^2) [-2.5, 2.5, 18.5, 30]}

If you want to cover Complex values too use:

#abs(z) = sqrt(zbar(z))#

So:

#y = sqrt((x^2-9)bar((x^2-9)))+sqrt((x^4-16)bar((x^4-16)))+sqrt((x^6-1)bar((x^6-1)))#

Answer 2 · 2016-12-15T02:50:25

Four piecewise definitions can be given. See explanation..

graph{-x^2-x^4+x^6+24 [-80, 80, -40, 40]}

The given equation is the combined form, for the four piecewise

definitions

#y=-(x^2-9)-(x^4-16)-(x^6-1)#

#=-x^2-x^4-x^6+26, x in [-1, 1]#

#=-(x^2-9)-(x^4-16)+(x^6-1)#

#=-x^2-x^4+x^6+24, x in [-2, -1] or [1, 2]#

#=-(x^2-9)+(x^4-16)+(x^6-1)#

#=-x^2+x^4+x^6-8, x in [-3, -2] or [2, 3]#

#=(x^2-9)+(x^4-16)+(x^6-1)#

#=x^2+x^4+x^6-26, x<=-3 or x >=3#.

As I am unable to get the all-in-one graph, the first piece can be

seen as the central part of the inserted second graph, with zenith at

(0, 26), and ends at #(+-1, 23)#

The second is in the first graph, with ends at #(+-24) and (+-2,

68)#. Likewise, interested readers can secure the other two pieces.

The common points #( x=+-1, +-2, +-3 )# of the pieces are

( 2-tangent ) nodes of this continuous function.

graph{sqrt(-x^2-x^4-x^6+26)^2 [-80, 80, -40, 40]}