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**Jacobpm****Guest**

Which of the following describes the behavior of y = cubicroot(x + 2) at x = -2

(A) differentiable

(B) corner

(C) cusp

(D) vertical tangent

(E) discontinuity

well i graphed the function, and i'm not sure.. i know for sure it isn't E... because f(-2) = 0... it has a value.. uhmm.. as for the rest i'm not sure.. i don't even know what a corner and a cusp is.

**krassi_holmz****Real Member**- Registered: 2005-12-02
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I don't know what is corner, but I think cusp is something like this:

I think (D)

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**Jacobpm****Guest**

like what? your picture didn't show up

**krassi_holmz****Real Member**- Registered: 2005-12-02
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Plot:

*Last edited by krassi_holmz (2006-01-02 12:23:24)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Now it's better.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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For uploading an image direct to the forum use "post reply" or when you've written your quick post, simply edit it. There you can specify the number of images and the path to be uploaded.

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**Jacobpm****Guest**

ok, so my grpah doesn't look like that.. so it doesn't have a cusp, correct?

so, hmm... we aren't sure about A, we aren't sure about B, we know it can't be C, we know it is D, and we know it can't be E..

So we're left with Discussions about A and B..

anything to add on those two?

**Jacobpm****Guest**

I just noticed that it isn't differentiable either...

So the only thing i'm unsure of is about the corner

**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yes, it isn't differenciable. And I don't know what is corner, too. Try at

http://www.mathworld.wolfram.com

*Last edited by krassi_holmz (2006-01-02 13:36:10)*

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**Ricky****Moderator**- Registered: 2005-12-04
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f(x) = abs(x), a corner exists at f(0). It is literally a corner.

As for the answer, it is D:

For a verticle tangent, the function's slope must approach infinity as it approaches the point.

So what we want is:

Multiplying this by

we get:Now x+2 approaches 0 as x approaches -2. But we know that f(x)^(1/3) > f(x) if f(x) < 1. So the numerator gets (relatively) larger and the denominator gets smaller as x approaches -2. Therefore the slope approaches infinity, and you have a verticle tangent.

*Last edited by Ricky (2006-01-02 15:36:15)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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D!

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**Ricky****Moderator**- Registered: 2005-12-04
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Oy, I completely missed a much more direct way to do the limit:

Since x+2 approaches zero as x approaches -2, x+2 is very close to 0 before it gets there, in other words, very small. (x+2)^(1/3) then also becomes increasingly small, and multiplying this by 3 has basically no effect as it will also become increasingly small. So 1 over this means it goes towards infinity.

*Last edited by Ricky (2006-01-03 05:22:19)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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It depends of which side you are limiting. If x < -2 and x --> -2 then x+2<0 and 1/(x+2) --> -oo.

If x>-2 and x --> -2 then x+2>0 and 1/(x+2) --> +oo.

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**Ricky****Moderator**- Registered: 2005-12-04
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True, but in this case, it doesn't matter. If it approaches negative infinity or infinity from either direction, it is still a vertical tangent.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yes, yes, just for exactude.

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