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Math Help - x-coordinates of tangent line

  1. #1
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    x-coordinates of tangent line

    I'm a little put off by the wording on this question, if anyone can offer some guidance on how to begin that would be great. also, sorry if i enter the symbols wrong, i'm new here. i don't know why the square root symbol isn't covering all of x^2 +7 , but it should.

    "find the x-coordinates of the points on the graph where the tangent line is horizontal"
    <br />
f(x) = \frac{x}{\sqrt(x^2 +7)}
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  2. #2
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    tangent line

    Hello,

    A horizontal tangentline means that:

    f'(x)=0

    All you have to do is:

    1) Find the derivative f'(x)

    2) Solve the equation f'(x)=0 with respect to x

    3) Done
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  3. #3
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    ok so when i'm doing the derivative i use chain rule to figure out the bottom part right?
    <br />
\frac {x}{\sqrt{x^2+7}}

    derivative of top = 1
    derivative of bottom using chain rule =

    u= x^2+7
    h(u) = \sqrt{u}

    \frac{d}{dx} \sqrt{x^2+7} = 2x*\sqrt{x^2+7} ??

    then use quotient rule for the rest?

    \frac {f\prime(x) * g(x) - f(x) * g\prime(x)}{g(x)^2}

    \frac{(1)(\sqrt{x^2+7}) - (x)(2x * \sqrt{x^2+7})}{\sqrt{x^2+7}^2} ???

    then would i just plug in 0 for x???

    thanks for your help!
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  4. #4
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    tangent line

    Hello,

    The derivative becomes:

    f'(x)=\frac{\sqrt{x^{2}+7}-x\cdot2x\cdot \frac{1}{2\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7})^{2}}=\  frac{\sqrt{x^{2}+7}- \frac{x^{2}}{\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7})^{2}  }=\frac{\sqrt{x^{2}+7}}{(\sqrt{x^{2}+7})^{2}} + \frac{\frac{x^{2}}{\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7  })^{2}}

    This could be simplified to:

    f'(x)=\frac{1}{\sqrt{x^{2}+7}} + \frac{x^{2}}{(\sqrt{x^{2}+7})^{\frac{3}{2}}}

    Now since the tangentline must be horizontal then:

    f'(x)=0

    Which means:

    \frac{1}{\sqrt{x^{2}+7}} + \frac{x^{2}}{(\sqrt{x^{2}+7})^{\frac{3}{2}}}=0

    In this equation you have to find x.

    Hope that helpes.
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