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Find the equation of a tangent line to y=f(x)

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent line to the curve at the given point.
    [tex]y = \sqrt{2x+1} , (4, 3)[/tex]


    2. Relevant equations
    a = 4
    [tex]M =\lim_{x \to a} \frac{f(x) - f(a)}{x-a}[/tex]

    or

    [tex]Mpq = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}[/tex]




    3. The attempt at a solution

    [tex]M =\lim_{x \to 4} \frac{f(x) - f(4)}{x-a}[/tex]

    [tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - \sqrt{2 * 4 + 1}}{x-4}[/tex]

    [tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4}[/tex]

    [tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4} * \frac{\sqrt{2x+1} + 3}{\sqrt{2x+1} + 3}[/tex]

    [tex]M =\lim_{x \to 4} \frac{2x - 3}{x-4(\sqrt{2x+1} + 3)}[/tex]

    At this point I get stuck... I'm not sure what I can do from here. did I go wrong some where?
     
  2. jcsd
  3. Mar 7, 2007 #2
    On the last step are you sure that you multiplied correctly? :P (check the numerator again)

    Do you see it?

    It should be 2x-8 since [tex](\sqrt{2x+1}-3) * (\sqrt{2x+1}+3) = 2x+1 -9 = 2x-8 [/tex]
     
    Last edited: Mar 7, 2007
  4. Mar 7, 2007 #3
    From that point you can divide by (x-4) quite easily resulting in a function for which you can directly substitute x=4 since the result is not in an 'indeterminate form' of [tex]\frac{0}{0}[/tex] or [tex]\frac{\infty}{\infty} [/tex]
     
    Last edited: Mar 7, 2007
  5. Mar 7, 2007 #4
    BAH! thanks :).
     
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