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Equation of Tangent Line

  1. Dec 7, 2004 #1
    Find the equation of the tangent line at x=2
    F(X)=2x+7x(e^(x-2))

    I know you put 2 in to find the y intercept and you take the derivative to find the slope. I get 2+(7e^(-2)x + 7e^-2)e^2 for my derivative and y=23x+18 as my answer and i know that is not correct.
     
  2. jcsd
  3. Dec 7, 2004 #2

    shmoe

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    You derivative looks off. Remember,

    [tex]\frac{d}{dx}e^{x-2}=e^{x-2}\frac{d}{dx}(x-2)=e^{x-2}(1)=e^{x-2}[/tex]

    and apply the product rule carefully!


    edit-ahh, Galileo is right, your slope is fine. Your posted derivative may be a casualty of ascii.
     
    Last edited: Dec 7, 2004
  4. Dec 7, 2004 #3

    Galileo

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    You got the slope right. So y=23x+b. Now use the fact that the line passes through the point (2,f(2)) to find b.
     
  5. Dec 7, 2004 #4

    dextercioby

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    The derivative should be:
    [tex] f'(x)=2+[7\exp({-2})](\exp{x}+x\exp{x) [/tex]
    Insert 2 to find the slope.It's 23.
    Compute f(2).It's 18.With the last 2 numbers,u should be able to find the equation for the tangent line.

    We're just like eagles when it comes to simple problems.Three almost identical posts at the same time...
     
    Last edited: Dec 7, 2004
  6. Dec 7, 2004 #5
    so was y=23x+18 correct?
     
  7. Dec 7, 2004 #6

    dextercioby

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    Obviously not,try to follow my post correctly.It's easy maths,after all.
     
  8. Dec 7, 2004 #7

    HallsofIvy

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    What is 23(2)+ 18?
     
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