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Homework Help: Find the equation of the tangent to the curve

  1. Mar 23, 2005 #1
    Find the equation of the tangent to the curve [tex]y=x^3-7x^2+14x-8[/tex] at the point where [tex]x = 1[/tex]. [tex]\text{Answer: }y = 3x -3[/tex]
    Find the x-coordinate of the point at which the tangent is parallel to the tangent at [tex]x = 1[/tex].

    I need help on the second part.
     
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  3. Mar 23, 2005 #2

    arildno

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    Well, let x* be the point you seek.
    What do you know of x*?
    Do you agree that we must have y'(x*)=3?
    That is, the derivatives of y(x) must be equal at x=1 and x=x*
     
  4. Mar 23, 2005 #3
    Yes.
    I think I get it. But I still don't know how to get the answer.
     
  5. Mar 23, 2005 #4

    arildno

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    Well, do you agree that what you need to solve is the equation (written with "x"):
    [tex]3=3x^{2}-14x+14[/tex]
    This can be rewritten as:
    [tex]3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0[/tex]
    What must then x* be?
     
  6. Mar 23, 2005 #5
    Oh! Thank you!
     
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