1. ## 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.
$
\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?

2. 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.

3. 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