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Calc Question: Finding Secant Slope?

  1. Jan 23, 2005 #1
    I wrote down the notes from class, but when I tried to do the homework, I am not even close to the right answers. The formula I wrote down is:
    [tex]\frac{-1}{(x)(x+h)}[/tex]
    Apparently that's wrong. Anyone know what it's supposed to be?
     
  2. jcsd
  3. Jan 23, 2005 #2
    Do you mean the derivative of [tex]f(x)=\sec x[/tex]? In that case it's [tex]\sec{x}\tan{x}[/tex].
     
  4. Jan 24, 2005 #3
    The point [tex]\mbox(p(1,\frac{1}{2}))[/tex] lies on the curve [tex]y=\frac{x}{1+x}[/tex].
    If [tex]Q[/tex] is the point [tex](x,\frac{x}{(1+x)})[/tex], use your calculator to find the slope of the secant line [tex]PQ[/tex] for the following value of [tex]x[/tex]: 0.5

    I get -2, but it's really .33333.
     
  5. Jan 24, 2005 #4

    dextercioby

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    The slope is simply
    [tex] a=\frac{y_{1}-y_{2}}{x_{1}-x_{2}} [/tex]

    and i get 1/3...

    Daniel.
     
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