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Math Help - Curve sketching problem

  1. #1
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    Curve sketching problem

    The problem is: f(x)=(9/x)+x+1 where x>0

    There are two asymptotes. I found x=0. But i'm stuck as I don't know how to get y=x+1 as an asymptote.

    I first solved f'(x) and made that equal zero which I got x=+/- 3. But because x>0, so x=3 ---> (3, 7)

    and I computed f''(x), plugged 3 in, the answer >0 so concave up and (3,7) is the minimum value.

    I know there is an asymptote at x=0, but what about the other one?

    Thanks!
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  2. #2
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    hi

    f(x)=\frac{x^2+x+9}{x}

    so obviously , the vertical asymtote is x=0

    the other one is an oblique/slant asymtote , divide x by x^2+x+9 (long division)
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  3. #3
    MHF Contributor

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    Quote Originally Posted by fabxx View Post
    The problem is: f(x)=(9/x)+x+1 where x>0

    There are two asymptotes. I found x=0. But i'm stuck as I don't know how to get y=x+1 as an asymptote.
    As x goes to \pm \infty, \frac{1}{x} goes to 0 so the graph is that of the rest of the formula, x+ 1.

    I first solved f'(x) and made that equal zero which I got x=+/- 3. But because x>0, so x=3 ---> (3, 7)

    and I computed f''(x), plugged 3 in, the answer >0 so concave up and (3,7) is the minimum value.

    I know there is an asymptote at x=0, but what about the other one?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

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