1. ## Critical points

$F(x) = \frac{p-4}{p^2+2}$

f(x) = p-4
f'(x) = -4
g(x) = $p^2 + 2$
g'(x) = 2p

$F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$

$F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

$

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.

2. Originally Posted by CFem
$F(x) = \frac{p-4}{p^2+2}$

f(x) = p-4
f'(x) = -4
g(x) = $p^2 + 2$
g'(x) = 2p

$F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$

$F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

$

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.
the derivative of $p-4$ is not $-4$

you also did the quotient rule incorrectly.

3. Originally Posted by CFem
$F(x) = \frac{p-4}{p^2+2}$

f(x) = p-4
f'(x) = -4
g(x) = $p^2 + 2$
g'(x) = 2p

$F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$

$F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

$

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.
For one thing, you have $F(x)=\frac{p-4}{p^2+2}$. With respect to $x$, 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 $F(x)=\frac{x-4}{x^2+2}$, we have:

$f(x)=x-4$
$f'(x)=1$
$g(x)=x^2+2$
$g'(x)=2x$

Therefore, $F'(x)=\frac{x^2+2-2x(x-4)}{(x^2+2)^2}=\frac{-x^2+8x+2}{(x^2+2)^2}$

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

$x=\frac{-8\pm\sqrt{64+8}}{-2}=4\pm3\sqrt{2}$

$F'(x)$ is defined everywhere, so there are no vertical tangents.

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

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

1) $F'(a)$ is undefined.
2) $F(a)$ is defined.

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

Take for example, $F(x)=\sqrt{x}$ and investigate the behavior of the tangent line at $x=0$.