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Math Help - Finding the "normal line"

  1. #1
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    Finding the "normal line"

    I understand in theory how to do this...it's just the particular problem that's throwing me.


    Question: Find the equation of the normal line to the graph of f(x) = e^x - 1 at the point where it crosses the x-axis.



    My Attempt:

    So f\prime(x) = e^x

    f(x) crosses the x-axis at pretty much x = 0 I think.

    so do I plug in 0 for x and solve in the original equation like this?
    <br />
f(x) = e^0 - 1<br />
      = 0

    Giving the point (0,0)

    then plug that into y = mx + b

    I get

    0 = e^x(0) + b
    so b = 0


    Does that mean the tangent line is just y= e^x(x)?

    Making the normal line  y = -\frac{1}{e^x}(x) ?





    Is that right? Or am I doing this completely wrong?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lysserloo View Post
    I understand in theory how to do this...it's just the particular problem that's throwing me.


    Question: Find the equation of the normal line to the graph of f(x) = e^x - 1 at the point where it crosses the x-axis.



    My Attempt:

    So f\prime(x) = e^x

    f(x) crosses the x-axis at pretty much x = 0 I think.

    so do I plug in 0 for x and solve in the original equation like this?
    <br />
f(x) = e^0 - 1<br />
= 0

    Giving the point (0,0)

    then plug that into y = mx + b

    I get

    0 = e^x(0) + b
    so b = 0


    Does that mean the tangent line is just y= e^x(x)?

    Making the normal line  y = -\frac{1}{e^x}(x) ?

    Is that right? Or am I doing this completely wrong?
    Note that f^{\prime}(0)=e^0=1.

    So y=x is the tangent line at that point, which then implies y=-x is the normal line at that point.
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