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Math Help - Concavity

  1. #1
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    Concavity

    f(x) = x^q(x+6)
    f'(x) = \frac{x+6}{5x^z} + x^!
    f''(x) = \frac{5x^z + \frac{x+6}{4x^q}}{25x^k}
    Note: I couldn't get fractional exponents to show up properly, so I used this key: q = \frac{1}{5} ; z = \frac{4}{5} ; k = \frac{8}{5}

    We need to find where f is concave up and down. There's two inflection points, one of which I've (correctly) found to be (0,0), which means that f is concave up at (-\infty,0) and from (s,\infty) and it's concave down from (0,s).

    But f''(x) never goes negative, and it's undefined at 0, so I'm not sure how to find s.
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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    f(x) = x^q(x+6)
    f'(x) = \frac{x+6}{5x^z} + x^!
    f''(x) = \frac{5x^z + \frac{x+6}{4x^q}}{25x^k}
    Note: I couldn't get fractional exponents to show up properly, so I used this key: q = \frac{1}{5} ; z = \frac{4}{5} ; k = \frac{8}{5}

    We need to find where f is concave up and down. There's two inflection points, one of which I've (correctly) found to be (0,0), which means that f is concave up at (-\infty,0) and from (s,\infty) and it's concave down from (0,s).

    But f''(x) never goes negative, and it's undefined at 0, so I'm not sure how to find s.

    Why to mess up with derivatives of first and second order of a product if you can enjoy yourself with a simple polynomial?? Write simply

    f(x)=x^{q+1}+6x^q

    Now evaluate the second derivative:

    f'(x)=(q+1)x^q + 6qx^{q-1}

    f''(x)=q(q+1)x^{q-1}+6q(q-1)x^{q-2}=qx^{q-2}[(q+1)x+6(q-1)]

    Where the second derivative is positive and where negative is easier now to see, I'm sure.

    Tonio
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