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Math Help - Critical Point Problem...

  1. #1
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    Critical Point Problem...

    Hi, this question is giving me a really hard time:



    With it, I'm supposed to: identify critical points, inflection points, asymptotes, where it is increasing, and where it is decreasing. I'm not really having a problem with the first three (I already found inflection points to be "none", as well as critical points), but I don't know how to find intervals of increase and decrease or intervals of concavity without having critical points or inflection points. I tried just looking at the first derivative, which I got to be 25/(x+6)^2) and reasoning that an x value above -6 would lead to increase, and below -6 would lead to decrease. This didn't work (although it did get me the vertical asymptote x=-6).
    I'd really appreciate help on this question. Thanks!
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  2. #2
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    Quote Originally Posted by fleabass123 View Post
    Hi, this question is giving me a really hard time:



    With it, I'm supposed to: identify critical points, inflection points, asymptotes, where it is increasing, and where it is decreasing. I'm not really having a problem with the first three (I already found inflection points to be "none", as well as critical points), but I don't know how to find intervals of increase and decrease or intervals of concavity without having critical points or inflection points. I tried just looking at the first derivative, which I got to be 25/(x+6)^2) and reasoning that an x value above -6 would lead to increase, and below -6 would lead to decrease. This didn't work (although it did get me the vertical asymptote x=-6).
    I'd really appreciate help on this question. Thanks!
    The graph is increasing for f'(x)>0
    The graph is decreasing for f'(x)<0

    f'(x)=\frac{(x+6)(3)-(3x-7)(1)}{(x+6)^2}=\frac{3x+18-3x+7}{(x+6)^2}=\frac{25}{(x+6)^2}

    This is always positive, hence the graph is always increasing, never decreasing, but x=-6 is not in the domain.
    There is a discontinuity at x=-6 and even though f(x) is extremely negative just to the right of x=-6, and is well below f(x) to the left of x=-6,
    the graph is never decreasing.
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  3. #3
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    Thanks for that! I'm also having the same problem with concave vs. convex. Without inflection points I don't know how to tell. Is there a method to do this?
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  4. #4
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    Yes,

    the 2nd derivative will give you this information.

    f''(x)=\frac{(x+6)^2(0)-25(2)(x+6)(1)}{(x+6)^4}=\frac{-50(x+6)}{(x+6)^4}

    This is negative for x>-6. This means the tangent slope is decreasing,
    causing the curve to be convex.

    The 2nd derivative is positive for x<-6.
    The tangent slope is increasing, meaning the curve is concave.
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  5. #5
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    Ahh, I was thinking I needed some kind of special formula but it turns out I just needed common sense...
    Thanks a lot for the help!
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