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Math Help - What value of x is a tangent parallel to the secant

  1. #1
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    What value of x is a tangent parallel to the secant

    Good evening folks.

    f(x) = 3^x - x^3
    For what value of x, is the tangent to the curve parallel to the secant through (0,1) and (3,0)?

    Do I find the slop between those 2 points and set it equal to the derivative of the original function?
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    Quote Originally Posted by alny View Post
    Good evening folks.

    f(x) = 3^x - x^3
    For what value of x, is the tangent to the curve parallel to the secant through (0,1) and (3,0)?

    Do I find the slop between those 2 points and set it equal to the derivative of the original function?
    Yes that's exactly what you do. Then solve for x.
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    Quote Originally Posted by Prove It View Post
    Yes that's exactly what you do. Then solve for x.
    Wow! Is it really? So if I do that it will turn into a logarithmic equation, correct? And solving that will get me my x?
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    Quote Originally Posted by alny View Post
    Wow! Is it really? So if I do that it will turn into a logarithmic equation, correct? And solving that will get me my x?
    Yes.

    Just remember though that you will need to use the chain rule to differentiate 3^x = e^{\ln{3^x}} = e^{x\ln{3}}.
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    3^x(ln3)-3x^2 = -1/3 ?
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    Quote Originally Posted by alny View Post
    3^x(ln3)-3x^2 = -1/3 ?
    Correct. Now solve for x if you can.
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    Quote Originally Posted by Prove It View Post
    Correct. Now solve for x if you can.
    This is as far as I got and i lost confidence that i was doing the right thing..

    3^x(ln3)-3x^2=-1/3 ... multiplied both sides by 3 to get rid of the fraction

    9^x(ln3) - 9x^2 = -1

    log(9^x(ln3) + 1) = log(9x^2)

    xlog9(ln3) + log(1) = 2log9x

    did I royally screw up somewhere?
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  8. #8
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    Continuing on..

    xlog9(ln3)-2log9x=log1
    x(log9ln(3) -2log9) = log 1
    x = log 1 / log9ln(3)-2log9
    x = 0 ??????
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  9. #9
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    Quote Originally Posted by alny View Post
    log(9^x(ln3) + 1) = log(9x^2)

    xlog9(ln3) + log(1) = 2log9x
    I don't think you can split a log up like that. Log(a+b) isn't Log(a) + Log(b)

    By the way, here are the answers for your eqaution: solve 3^x ln(3) -3x^2 = (-1/3) for x - Wolfram|Alpha
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