1. ## 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?

2. 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.

3. Originally Posted by Craka
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