# Critical points

• Oct 12th 2009, 03:53 PM
CFem
Critical points
$\displaystyle F(x) = \frac{p-4}{p^2+2}$

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

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

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

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.
• Oct 12th 2009, 06:01 PM
skeeter
Quote:

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

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

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

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

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 $\displaystyle p-4$ is not $\displaystyle -4$

you also did the quotient rule incorrectly.
• Oct 12th 2009, 06:07 PM
redsoxfan325
Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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