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Math Help - tangent line curve

  1. #1
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    tangent line curve

    Hi guys.

    equation of tangent line to the curve y = 3x^2 + e^{1-x} when x=1

    i tried and got 0

    can you show me how to solve with working please.
    thank you!
    Last edited by tankertert; April 17th 2012 at 03:02 AM.
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  2. #2
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    Re: tangent line curve

    dy/dx=6x-1 So when x=1 gradient of tangent=5 Also when x=1 y=2+e

    So you want a straight line with gradient 5 through the point (1,2+e)
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  3. #3
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    Re: tangent line curve

    Quote Originally Posted by tankertert View Post

    equation of tangent line to the curve y = 3x^2 + e^1-x when x=1
    I suspect OP meant y = 3x^2 + e^{1-x}.

    If so OP remember to write it like this e^{1-x}.
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  4. #4
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    Re: tangent line curve

    Quote Originally Posted by a tutor View Post
    I suspect OP meant y = 3x^2 + e^{1-x}.

    If so OP remember to write it like this e^{1-x}.
    sorry yes that what i mena
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  5. #5
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    Re: tangent line curve

    So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5
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  6. #6
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    Re: tangent line curve

    Quote Originally Posted by biffboy View Post
    So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5
    but how did you get that
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