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Thread: linear approximation

  1. #1
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    linear approximation

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

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

    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 $\displaystyle \frac{1}{x}$? o_O

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

    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 $\displaystyle f(x) = \frac{1}{x}$ at $\displaystyle x = 2$.
    If $\displaystyle \Delta x = 0.1$ then $\displaystyle \Delta y = ?$

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

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


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

    so:

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

    RonL
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