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Equation of a tangent

  1. Apr 24, 2007 #1
    I have a function.

    I am asked to find the equation of the tangent to the curve at (3, f(3))

    I think I am just confused of notation, but what is it asking here? To find the points, don't you put in the x value into the first derivative and this gets the y value, and use the formula for an equation of a line? What are those coordinates for then?
  2. jcsd
  3. Apr 24, 2007 #2


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    Well, you're supposed to find the equation of the tangent at that very point.
  4. Apr 24, 2007 #3
    So I already have my x and y, so just use y-y1=m(x-x1)????
  5. Apr 24, 2007 #4


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    If you meant y - f(c) = f'(c)(x - c), then yes. (c = 3)
  6. Apr 24, 2007 #5
    As you might know f'(c)=tg@, so the derivative of the curve at the point (c,f(c)) is actually the orientation coeficient of the tagent drawn at that point.

    So this is actually what radou said, just giving u some more hints.
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