I don't know what is corner, but I think cusp is something like this:
I think (D)

Plot:

Last edited by krassi_holmz (2006-01-03 11:23:24)

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

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

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 (2006-01-03 14:36:15)

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 (2006-01-04 04:22:19)

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.