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Tangent to y=e^x

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find an equation for a line that is tangent to the graph of y=ex and goes through the origin.


    2. Relevant equations



    3. The attempt at a solution
    y'=ex

    That's about all I can think of. I don't know how to make the tangent line go through the origin. Can someone lead me in the right direction?
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    Ray Vickson

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    At any point (x,y) on the graph, how would you compute the slope of the tangent line?

    RGV
     
  4. Dec 1, 2011 #3
    Find the derivative at that point.
     
  5. Dec 2, 2011 #4

    HallsofIvy

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    And you have already said that the derivative is again [itex]e^x[/itex].

    So the tangent line to [itex]y= e^x[/itex] at [itex]x= x_0[/itex] would be [itex]y= e^{x_0}(x- x_0)+ e^{x_0}[/itex]. Now, what must [itex]x_0[/b] be so that goes through (0, 0)?
     
  6. Dec 2, 2011 #5
    0? Not really sure..
     
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