Question about function y = x^2 - abs(x-4)

[CrazCo]

y = x^2 - abs(x-4)

 

at x = 4 how do i tell if it has a discontinuity, cusp, vertical tangent or corner using algebra

 

 

im soooo lost.

[theoriginal169]

draw a graph

[CrazCo]

we have to use algebra

[theoriginal169]

if u draw a graph you can see your answer

[CrazCo]

i know, but i think im doing it wrong because it looks completely continuous but apparently it has a corner meaning it has a different limit from each side at x=4

[timo]

Shouldn't it run to the same corner from both sides? I think you are mixing up continuous and differentiable.

[CrazCo]

im lost..

[mooeypoo]

CrazCo,

 

As with any question in math, try to break it down to smaller chunks, so it gets less scary and easier to handle. Start piece by piece.

 

Point of discontinuity means that at some point the function is undefined. Is there such a point?

When x=4, you simply have a zero inside the absolute value, and your function remains f(4)=16. That's not a discontinuity. Is there any other point of discontinuity?

 

For vertical tangent you need to calculate the tangent first, then equate it to zero -- it is the same as finding maxima and minima... do you know how to calculate a tangent line?

 

I don't know if it's a language barrier, but I have no idea what Cusp or Corner mean in the context of algebra, so I can't help you there.

 

 

 

 

~moo

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Merged post follows:

Consecutive posts merged

BTW, if a "corner" means a point where the function "Breaks" sharply (hence, not continuous) you should be able to do that with tangents as well.

 

Which means your next step is to calculate the general tangent line.

[CrazCo]

cusp = limit is a different finite number from both sides

corner = limit is a infinity from one side and negative infinity from the other

 

i think it's a cusp because when you derive it you get

 

y = x^2 - |x-4|

 

y=x^2 + x - 4

y= x^2 -x + 4

 

y' = 2x+1

y'= 2x-1

 

as it approaches 4

 

y' = 9

y' = 7