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?
2 Answers
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.
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.
