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Math Help - finding critical points

  1. #1
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    finding critical points

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

    <br />
\begin{array}{l}<br />
 f(b) = \frac{{b + 3}}{{b^2  + 2b}} \\ <br />
 f'(b) = \frac{{ - b^2  - 6b - 6}}{{b^2 (b + 2)^2 }} \\ <br />
 f''(b) = \frac{{2(b^3  + 9b^2  + 18b + 12)}}{{b^3 (b + 2)^3 }} \\ <br />
 \end{array}<br />
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Craka View Post
    how does one go about finding the critical points to this function, given its first and 2nd derivative.

    <br />
\begin{array}{l}<br />
 f(b) = \frac{{b + 3}}{{b^2  + 2b}} \\ <br />
 f'(b) = \frac{{ - b^2  - 6b - 6}}{{b^2 (b + 2)^2 }} \\ <br />
 f''(b) = \frac{{2(b^3  + 9b^2  + 18b + 12)}}{{b^3 (b + 2)^3 }} \\ <br />
 \end{array}<br />
    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
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  3. #3
    Moo
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    Hi Chris,
    Quote Originally Posted by Chris L T521 View Post
    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 ?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    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
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