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Math Help - UGGGH! How many points of inflection does f(x) have?

  1. #1
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    UGGGH! How many points of inflection does f(x) have?

    f(x)=(\frac {\pi}{3})^{x^3-8}

    Don't laugh at me, but I tried to do this algebraically, which was TERRIBLE.

    I need to show work, whatever that means.

    a. none
    b. one
    c. two
    c. three
    e. Infinitely many
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JOhkonut View Post
    f(x)=(\frac {\pi}{3})^{x^3-8}

    Don't laugh at me, but I tried to do this algebraically, which was TERRIBLE.

    I need to show work, whatever that means.

    a. none
    b. one
    c. two
    c. three
    e. Infinitely many
    hint: f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}

    and

    f''(x) = 6x \cdot \ln \frac {\pi}3 \cdot \bigg( \frac {\pi}3 \bigg)^{x^3 - 8} + (3x^2)^2 \bigg( \ln \frac {\pi}3 \bigg)^2 \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    hint: f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}

    and

    f''(x) = 6x \cdot \ln \frac {\pi}3 \cdot \bigg( \frac {\pi}3 \bigg)^{x^3 - 8} + (3x^2)^2 \bigg( \ln \frac {\pi}3 \bigg)^2 \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}
    I already tried that, but at the end, how can I be sure to get all the zeros with my graphing calculator? I can't find the roots algebraically either...
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  4. #4
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    wait, thats not the same second derivative as I got... I don't see how you got that second term. Mine was more complicated.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JOhkonut View Post
    wait, thats not the same second derivative as I got... I don't see how you got that second term. Mine was more complicated.
    Oh? what did you get? (note that we have a product of functions-- 3x^2 is a function as well as \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}, hence we need the product rule, which yields two terms)
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  6. #6
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    oh i see now, I treated  \ln \frac {\pi}{3} as a term rather than a constant.

    This makes things much easier, thanks!
    Last edited by JOhkonut; August 28th 2008 at 06:55 PM.
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  7. #7
    Super Member 11rdc11's Avatar
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    u = 3x^2
    du = 6x
    v =(\frac {\pi}{3})^{x^3-8}
    dv = 3x^2 \bigg( \frac {\pi}3\bigg)^{x^3 - 8}


    ln \frac{\pi}{3}[(du)(v) + (dv)(u)]

    oops too slow again sorry the latex slowed me down
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