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Absolute Value Functions

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Collin Li:
Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

If you differentiate a modulus function, and the limits do not agree, then open circles will reside on the x-coordinate of the cusp.

joechan521:

--- Quote from: "asa.hoshi" ---what about absolute functions that are like this:
y= |x+1| - |x+2|
--- End quote ---



ok, this graph is kinda hard to draw by hand, but it is possible.

for x>-1 both x+1 and x+2 is positive
therefore
y=(x+1)-(x+2)
y=-1 for values  x>-1

for    -2<x<-1   we can see that x+1 is negative, and x+2 is positive
y=-(x+1)-(x+2)
y=-2x-3
x+2>0 x+1<0    -1>x>-2

for    x<-2
both x+1 and x+2 is negative
y=-(x+1)+(x+2)
y=1.

therefore the whole graph would have 3 sections
y=1 for  x<-2
y=-2x-3 for -2<=x<-1
y=-1 for -1<=x

and they will intesect at points -1and-2, so no need to worry open or close circle

Ahmad:

--- Quote from: "coblin" ---Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

--- End quote ---


Coblin is right.

Freitag:

--- Quote from: "Ahmad" ---
--- Quote from: "coblin" ---Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

--- End quote ---


Coblin is right.
--- End quote ---


Is he ever wrong? Damn your 80/80 for methods paper 2 Coblin..

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