f(x) = p-4

f'(x) = -4

g(x) =

g'(x) = 2p

I understand that a critical number is where F'(x) is either c or DNE.

But as far as I can tell, neither the bottom nor the top can be equal to 0, which I think neglects both cases.

Printable View

- Oct 12th 2009, 03:53 PMCFemCritical points

f(x) = p-4

f'(x) = -4

g(x) =

g'(x) = 2p

I understand that a critical number is where F'(x) is either c or DNE.

But as far as I can tell, neither the bottom nor the top can be equal to 0, which I think neglects both cases. - Oct 12th 2009, 06:01 PMskeeter
- Oct 12th 2009, 06:07 PMredsoxfan325
For one thing, you have . With respect to , this is a constant function (and therefore has no critical points). I'm not sure if that's what you meant, but with the case , we have:

Therefore,

Using the quadratic formula on the numerator, we get that has two critical points (with horizontal tangent lines) at:

is defined everywhere, so there are no vertical tangents.

----------------------

On a related note, it only makes sense to define a vertical tangent of at when:

1) is undefined.

2) is defined.

Otherwise, you'll just get the equation for the asymptote.

Take for example, and investigate the behavior of the tangent line at .