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Math Help - linear approximation y=

  1. #1
    Member
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    Jun 2008
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    175

    linear approximation y=

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

    then y=?

    they give the answer to be y =  - \frac{1}{4}x + 1

    I get this to be the answer
    <br />
\begin{array}{l}<br />
 y = f'(a)(x - a) + f(a) \\ <br />
  =  - \frac{1}{4}x - 2 + \frac{1}{2} \\ <br />
  =  - \frac{1}{4}x - 1.5 \\ <br />
 \end{array}<br />

    Have I done something wrong, and if so where?
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Yes, you've made an error; it's in your algebra:
    y=f'(a)(x-a)+f(a)
    y=-\frac{1}{4}(x-2)+\frac{1}{2}
    y=-\frac{1}{4}\cdot{x}-\left(-\frac{1}{4}\right)\cdot(2)+\frac{1}{2}
    y=-\frac{1}{4}\cdot{x}+\frac{1}{2}+\frac{1}{2}
    y=-\frac{1}{4}\cdot{x}+1

    --Kevin C.
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  3. #3
    Member
    Joined
    Jun 2008
    Posts
    175
    ah bugger. Thinking I should have had more sleep.
    Thanks
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