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Yes, yes, just for exactude.
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.
It depends of which side you are limiting. If x < -2 and x --> -2 then x+2<0 and 1/(x+2) --> -oo.
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.
f(x) = abs(x), a corner exists at f(0). It is literally a corner.
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.
Yes, it isn't differenciable. And I don't know what is corner, too. Try at
I just noticed that it isn't differentiable either...
ok, so my grpah doesn't look like that.. so it doesn't have a cusp, correct?
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Now it's better.
like what? your picture didn't show up
I don't know what is corner, but I think cusp is something like this:
Which of the following describes the behavior of y = cubicroot(x + 2) at x = -2