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Tangent Lines

  1. Sep 12, 2005 #1
    I need to find an equation for the tangent line to the graph of the function at the specified point, I had some trouble simplifing f(x) so that I can take the lim as h->0 which is the slope.

    f(x) = Sqrt(x)/5 at (4,(2/5))
    Last edited: Sep 12, 2005
  2. jcsd
  3. Sep 12, 2005 #2


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    Do you mean that you have to use the basic formula for derivative
    [tex]lim_{h->0}\frac{f(a+h)-f(a)}{h}[/tex] rather than derivative formulas?

    Try multiplying both numerator and denominator of
    by [tex]\sqrt{a+h}+\sqrt{a}[/tex].
  4. Sep 12, 2005 #3
    I did that and I got something like 1/2*Sqrt(a) by taking the lim as h->0, giving a slope of 1/4, but the answer is sopposed to be y=(1/20)*x+(1/5) can you please show me how they got this ?

    By the way is that factor you posted called a conjugate ?
    Last edited: Sep 12, 2005
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