1. ## finding critical points

how does one go about finding the critical points to this function, given its first and 2nd derivative.

$
\begin{array}{l}
f(b) = \frac{{b + 3}}{{b^2 + 2b}} \\
f'(b) = \frac{{ - b^2 - 6b - 6}}{{b^2 (b + 2)^2 }} \\
f''(b) = \frac{{2(b^3 + 9b^2 + 18b + 12)}}{{b^3 (b + 2)^3 }} \\
\end{array}
$

2. Originally Posted by Craka
how does one go about finding the critical points to this function, given its first and 2nd derivative.

$
\begin{array}{l}
f(b) = \frac{{b + 3}}{{b^2 + 2b}} \\
f'(b) = \frac{{ - b^2 - 6b - 6}}{{b^2 (b + 2)^2 }} \\
f''(b) = \frac{{2(b^3 + 9b^2 + 18b + 12)}}{{b^3 (b + 2)^3 }} \\
\end{array}
$
To find all stationary points, set the first derivative equal to zero:

$\frac{{ - b^2 - 6b - 6}}{{b^2 (b + 2)^2 }}=0$

This implies that $-b^2-6b-6=0\implies b^2+6b+6\neq0$

To find the remaining critical points, look for a value of b that causes the first derivative to be undefined:

In other words, where $\frac{{ - b^2 - 6b - 6}}{{b^2 (b + 2)^2 }}=\infty$

This implies that $b^2(b+2)^2=0\implies b=\dots~or~b=\dots$

These two b values are the only critical points.

I hope this helps!

--Chris

3. Hi Chris,
Originally Posted by Chris L T521
This implies that $-b^2-6b-6=0\implies b^2+6b+6\neq0$
My neurons may have had some severe damage (don't laugh, it's highly possible lol !), but why is it $\neq 0$ ?

$x=-3\pm \sqrt{3}$ is solution, isn't it ?

4. Originally Posted by Moo
Hi Chris,

My neurons may have had some severe damage (don't laugh, it's highly possible lol !), but why is it $\neq 0$ ?

$x=-3\pm \sqrt{3}$ is solution, isn't it ?
Shoot....>_<

Thanks for catching that...

--Chris