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Math Help - linear approximation

  1. #1
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    linear approximation

    problem follows, "let y be the linearisation of the function f(x) = \frac{1}{x} at x = 2.
    If \Delta x = 0.1 then \Delta y = ?

    This is what I get.
    <br />
\begin{array}{l}<br />
 \Delta y = f(x + \Delta x) - f(x) \\ <br />
  = f(2 + 0.1) - f(2) \\ <br />
  = \frac{1}{{2.1}} - \frac{1}{2} \approx  - 0.0238 \\ <br />
 \end{array}<br />

    However answer is give as being -0.025. Rounding my above answer only give -0.024
    Have I done something wrong?


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  2. #2
    Super Member
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    Linear approximation for \frac{1}{x}? o_O

    For linear approximation, you need to find the equation:
    f(x) = f'(a)(x-a) + f(a)

    This function is an approximate to the original function for x near a. I'm sorry for being of little help but I don't know what are you trying to do.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Craka View Post
    problem follows, "let y be the linearisation of the function f(x) = \frac{1}{x} at x = 2.
    If \Delta x = 0.1 then \Delta y = ?

    This is what I get.
    <br />
\begin{array}{l}<br />
\Delta y = f(x + \Delta x) - f(x) \\ <br />
= f(2 + 0.1) - f(2) \\ <br />
= \frac{1}{{2.1}} - \frac{1}{2} \approx - 0.0238 \\ <br />
\end{array}<br />

    However answer is give as being -0.025. Rounding my above answer only give -0.024
    Have I done something wrong?


    y(x+\delta)=f(x)+\delta f'(x)=\frac{1}{x}-\delta \frac{1}{x^2}

    so:

    y(2+0.1)=\frac{1}{2}-\frac{0.1}{4}=y(2)-0.025

    RonL
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