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Math Help - increasing on interval?

  1. #1
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    Smile increasing on interval?

    Given the below function and its 1st and 2nd derivative how do I go about testing if f(b) is increasing on the interval  - 3 - \sqrt 3  < b <  - 3 + \sqrt 3




    <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
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    Find the points at which the first derivative is zero, which we call critical points. Then plug in points that precedes or succeeds the critical points and check it's behavior (negative/positive). You can determine then whether your function is increasing on your interval.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Craka View Post
    Given the below function and its 1st and 2nd derivative how do I go about testing if f(b) is increasing on the interval  - 3 - \sqrt 3  < b <  - 3 + \sqrt 3




    <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 test if its increasing on this interval, plug a value within the interval into the first derivative. For the function to be increasing, f'(b)>0

    To show that its increasing only within this interval, pick points outside of the interval [one value b<-3-\sqrt{3} and another value b>-3+\sqrt{3}] and show that f'(b)<0 for these values.

    I'm not quite sure [at the moment] of how to apply the second derivative to this, but I think this may help you out enough.

    I hope this helps!

    --Chris
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