How do you find vertical tangents and cusps?

f(x)=(x+4)^-4/7 at c=4

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

Question

How do you find vertical tangents and cusps?

f(x)=(x+4)^-4/7 at c=4

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

anonymous · Answer

Specifically for these problems

(a) f(x)=(x+4)−4/7 at c=−4.
(b) f(x)=7(x−2)3/7 at c=2.

I have the answers, but I don't know how to approach the method.

zepdrix · Answer

So like, you know how you set your first derivative equal to zero....
and solving for x gives you `horizontal asymptotes` usually, yes?

zepdrix · Answer

Where your first derivative is undefined is usually where you're running into these vertical asymptotes or cusps.

zepdrix · Answer

Do you understand how to find f'(x) for the first problem?

zepdrix · Answer

This first question seems poorly written.
The function doesn't even exist at x=-4, how could have a vertical tangent there?
Hmm that's strange :(

jhannybean · Answer

If the limits approach two different values, it usually means there is a cusp. :P

zepdrix · Answer

The second one makes sense though :)$$\Large\rm f(x)=7(x-2)^{3/7}$$Then,$$\Large\rm f'(x)=\frac{3}{(x-2)^{4/7}}$$Looking for tangents,$$\Large\rm 0=\frac{3}{(x-2)^{4/7}}$$We can see that the first derivative doesn't exist at (x-2)^{4/7}=0
(because we can't divide by 0 in the land of math).

Solving for x gives us x=2.

zepdrix · Answer

So what's your question exactly SizzlinBeef? 0_o
Just the whole process confusing?