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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 x->0 which is the slope.

    f(x) = Sqrt(x)/5 at (4,(2/5))
  2. jcsd
  3. Sep 12, 2005 #2
    *Simply find the derivative of the curve at that point, and use point-slope to find the line.
    [tex] f\left( x \right) = \frac{\sqrt x}{5} \Rightarrow f\,'\left( x \right) = \frac{1}{{10\sqrt x }} [/tex]

    *To find the slope of the tangent line, simply calculate [itex] f\,' ( 4 ) [/itex]:
    [tex] f\,'\left( 4 \right) = \frac{1}{{20}} [/tex]

    *Next, just use point-slope to represent the tangent line, which I'll call [itex] y [/itex]:
    [tex] \frac{{x - 4}}{{20}} = y - \frac{2}{5} [/tex]

    And it's algebra from there :smile:
    Last edited: Sep 12, 2005
  4. Sep 13, 2005 #3


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    Are you saying that you are required to use the basic formula for the derivative: [itex] lim_{h->0}\frac{f(a+h)-f(a)}{h}[/itex] rather than the more specific formula (derived from that) that bomba923 used?

    If so try multiplying both numerator and denominator of
    [tex]\sqrt{a+h}+ \sqrt{a}[/tex].
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