# Function has vertical tangent or vertical cusp?

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?

• Your $\ f(x) \$ is a polynomial function: since the derivative function of a polynomial function is always defined, it does not have vertical tangents or cusps. – colormegone Jul 12 '14 at 22:37
• You have shown that $f(x)$ has an extreme point at $c = -2$. – Code-Guru Jul 13 '14 at 0:10
• A vertical tangent or cusp would appear where the slope diverges. You have shown that the slope at that point remains finite. Thus there is no vertical tangent. – zahbaz Feb 8 '16 at 21:30

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