linear approximation

**Craka** · September 10th 2008, 05:10 AM

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.
$ \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?

**Chop Suey** · September 10th 2008, 05:20 AM

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.

**CaptainBlack** · September 10th 2008, 05:35 AM

Originally Posted by Craka

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.
$ \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?

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

so:

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

RonL

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