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Calc tangent problem

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f(x)=a(7-x^2)
    find in terms of a, the equations of the lines tangent to these curves at x=-1.

    2. Relevant equations
    ?

    3. The attempt at a solution
    So I took the derivative of f(x).
    f'(x)=a(-2x)+7-x^2
    then i plugged in -1 into f'(x)
    and i got f'(-1)=2ax+7+1=2ax+8
    but the answer to the problem is y=2ax+8a.
    I dont know where I went wrong.
     
  2. jcsd
  3. Apr 20, 2009 #2

    Cyosis

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    It went wrong at the part where you took the derivative. Write out the brackets and then ask yourself what's [tex]\frac{d7a}{dx}, \frac{-dx^2}{dx}[/tex].
     
  4. Apr 20, 2009 #3

    dx

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    The derivative of a(7 - x²) is -2ax.

    The slope of a line is not enough to determine the equation of the line. If you have the slope, and also its height at a point b, call it c, then the equation of the line will be:

    y = c + (x - b)(slope)

    Do you recognize this?
     
  5. Apr 20, 2009 #4
    ok
    so f'(x)=-2ax which is the slope.
    then to get the set of points
    you know x=-1,
    so f(-1)=8a.
    but if you used point slope format
    wouldnt you get y-8a=-2ax(x+1) ?
     
  6. Apr 20, 2009 #5

    dx

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    Yes, except you have to replace -2ax in your equation with -2a(-1) since you want the slope at -1.
     
  7. Apr 20, 2009 #6
    o ok.
    but wouldnt you still get y-8a=2a(x+1)
    which is y=2ax+10a?
     
  8. Apr 20, 2009 #7

    dx

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    You calculated f(-1) wrong. It should be 6a not 8a.
     
  9. Apr 25, 2009 #8
  10. Apr 25, 2009 #9

    dx

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    No problem. :smile:
     
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