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Determine whether or not the graph of the function has a vertical tangent or vertical cusp at the indicated point c.

$f(x) = (x+2)^7/3$
$c=-2$

I took the first derivative and chain rule and that got me.

$ f'(x) = 7/3(x+2)^4/3*1$

Then I plugged in c

$f'(-2) = 7/3(-2+2)^4/3 = 0$

Then I put that into a line graph.

___0______

Then I got

$f(-1) =1^7/3$

$f(1) = 2^7/3$

Thus this is a vertical cusp. Is this right or wrong?

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2 Answers

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The definition of a vertical cusp is that the one-sided limits of the derivative approach opposite $ \pm \infty $: positive infinity on one side and negative infinity on the other side. A vertical tangent has the one-sided limits of the derivative equal to the same sign of infinity. As a result, the derivative at the relevant point is undefined in both the cusp and the vertical tangent.

You have a case where the derivative exists, as you showed in your question. Therefore, it is neither a cusp nor a vertical tangent.

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At $x=2$, the tangent line is horizontal, since the derivative at that point is zero.

A vertical tangent means that the derivative at that point approches infinity, since the slope is infinitely large.

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