linear approximation y=

**Craka** · September 10th 2008, 04:10 PM

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
$ \begin{array}{l} y = f'(a)(x - a) + f(a) \\ = - \frac{1}{4}x - 2 + \frac{1}{2} \\ = - \frac{1}{4}x - 1.5 \\ \end{array} $

Have I done something wrong, and if so where?

**TwistedOne151** · September 10th 2008, 04:55 PM

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.

**Craka** · September 10th 2008, 05:01 PM

ah bugger. Thinking I should have had more sleep.

Thanks

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