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Thread: Concavity

  1. #1
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    Concavity

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

    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 $\displaystyle (-\infty,0)$ and from $\displaystyle (s,\infty)$ and it's concave down from $\displaystyle (0,s)$.

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

    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 $\displaystyle (-\infty,0)$ and from $\displaystyle (s,\infty)$ and it's concave down from $\displaystyle (0,s)$.

    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

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

    Now evaluate the second derivative:

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

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

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

    Tonio
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