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Math Help - Trouble Finding Tangent Lines

  1. #1
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    Trouble Finding Tangent Lines

    The graph of y = 6(3x^2)^x has two horizontal tangent lines. Find the equations for both of them.
    Alright, so my approach will be to find the derivative of the function using logarithmic differentiation or b^x = e^\ln b^x. Next, I'll set this derivative equal to zero and solve to obtain the tangent lines, since a tangent line occurs when f'(x) = 0. Unfortunately, both attempts have been unsuccessful.

    Logarithmic Differentiation

    y = 6(3x^2)^x

    \ln y = \ln [6(3x^2)^x]

    \frac{dy}{dx} \cdot \frac{1}{y} = \ln 6 + \ln (3x^2)^x

    \frac{dy}{dx} \cdot \frac{1}{y} = 0 + x\ln 3x^2

    \frac{dy}{dx} \cdot \frac{1}{y} = (1 \cdot \ln 3x^2) + x(\frac{1}{3x^2} \cdot 6x)

    \frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + \frac{6x^2}{3x^2}

    \frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + 2

    \frac{dy}{dx} = 6(3x^2)^x[\ln 3x^2 + 2]

    Solve y' For 0 To Find Horizontal Tangents Of Original Function

    \ln(3x^2) + 2 = 0

    \ln(3x^2) = -2

    e^{\ln(3x^2)} = e^{-2}

     3x^2 = e^{-2}

    x^2 = \frac{e^{-2}}{3}

    \sqrt{x^2} = \frac{\sqrt{e^{-2}}}{\sqrt{3}}

    x = \pm \frac{1}{e\sqrt{3}}

    Evaluate Original Function At x

    y = \ln(3x^2)^x

    f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{e\sqrt{3}})^2)^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{3e^2}))^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(\frac{3}{3e^2})^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(\frac{1}{e^2})^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(e^{-2})^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(e^{\frac{-2}{1} \cdot \frac{\sqrt{3}}{3e}})

    y = 6e^{\frac{-2\sqrt3}{3e}}

    (Substitute -\frac{1}{e\sqrt{3}} to get the other solution)

    The answer is said to be:

    y = 6e^{\frac{-2\sqrt3}{3e}} and y = 6e^{\frac{2\sqrt3}{3e}}

    Which confirms that the solution is correct.

    Thanks a lot for the help!
    Last edited by Algebrah; October 17th 2011 at 07:56 PM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Trouble Finding Tangent Lines

    You have calculated the first derivative f'(x) correctly. Indeed, if you have to find the horizontal tangent lines (their slope =0) to the curve therefore you have to solve:
    6(3x^2)^x\cdot \ln[3x^2+2]=0 \Leftrightarrow 6(3x^2)^x=0 \ \mbox{or} \ \ln(3x^2)+2=0
    There're no solutions for 6(3x^2)^x=0, so you have to solve \ln(3x^2)+2=0, can you do that?

    Afterwards you have found a point on the curve where you get a horizontal tangent line, so you only have to find the y- coordinate afterwards by substituting the x- value in the equation of the curve.
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  3. #3
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    Re: Trouble Finding Tangent Lines

    It appears that my logarithmic and exponential function manipulation skills are egregious. I need to learn to get comfortable with these. xD

    Here's what I tried:

    \ln(3x^2) + 2 = 0

    \ln(3x^2) = -2

    e^{\ln(3x^2)} = e^{-2}

     3x^2 = e^{-2}

    x^2 = \frac{e^{-2}}{3}

    \sqrt{x^2} = \frac{\sqrt{e^{-2}}}{\sqrt{3}}

    x = \pm \frac{1}{e\sqrt{3}}

    I'm relatively sure that I made an error up this point.

    y = \ln(3x^2)^x

    f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{e\sqrt{3}})^2)^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{3e^2}))^{\frac{1}{e\sqrt{3}}}

    f(\frac{1}{e\sqrt{3}}) = 6(\frac{3}{3e^2})^{\frac{1}{e\sqrt{3}}}
    Last edited by Algebrah; October 17th 2011 at 06:15 PM.
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  4. #4
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    Re: Trouble Finding Tangent Lines

    \ln(3x^2) + 2 = 0

    \ln(3x^2) = -2

    3x^2 = \frac{1}{e^2}

    x = \pm \frac{1}{\sqrt{3} \cdot e}


    note that \sqrt{x^2} = |x|
    Last edited by Ackbeet; October 17th 2011 at 06:06 PM.
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  5. #5
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    Re: Trouble Finding Tangent Lines

    Sorry about that.

    The part that's confusing me is how they're getting

    y = 6e^{\frac{-2\sqrt3}{3e}} and y = 6e^{\frac{2\sqrt3}{3e}}

    as an answer. I've been working on this one for a while now and it's the only one I can't get.

    Is that answer a byproduct of using the b^x = e^\ln b^x method before substituting the x values into the original function to find the y values?

    Normally, I wouldn't be so aggressive in my inquiries, but multiple people are having the same problem with this one in my class. :x
    Last edited by Algebrah; October 17th 2011 at 06:50 PM.
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  6. #6
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    Re: Trouble Finding Tangent Lines

    Quote Originally Posted by Algebrah View Post
    Sorry about that.

    The part that's confusing me is how they're getting

    y = 2e^{\frac{-\sqrt6}{3e}} and y = 2e^{\frac{\sqrt6}{3e}}

    as an answer.
    those solutions do not agree with the graph of the original function y = 6(3x^2)^x

    either the original function or the solutions for the horizontal tangent lines are incorrect.
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  7. #7
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    Re: Trouble Finding Tangent Lines

    Here's the problem:


    EDIT: Alright, so it turns out that I copied down the answer to a similar problem.

    I finished it. Always a good idea to make sure you're copying down the right thing. Haha.

    Thanks everyone!
    Last edited by Algebrah; October 17th 2011 at 07:41 PM.
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