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

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.

]]>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.

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

]]>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?

]]>I think (D)]]>

(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.

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