
linear approximation y=
problem follows, "let y be the linearisation of the function $\displaystyle f(x) = \frac{1}{x}$ at $\displaystyle x = 2$.
then y=?
they give the answer to be $\displaystyle y =  \frac{1}{4}x + 1$
I get this to be the answer
$\displaystyle
\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?

Yes, you've made an error; it's in your algebra:
$\displaystyle y=f'(a)(xa)+f(a)$
$\displaystyle y=\frac{1}{4}(x2)+\frac{1}{2}$
$\displaystyle y=\frac{1}{4}\cdot{x}\left(\frac{1}{4}\right)\cdot(2)+\frac{1}{2}$
$\displaystyle y=\frac{1}{4}\cdot{x}+\frac{1}{2}+\frac{1}{2}$
$\displaystyle y=\frac{1}{4}\cdot{x}+1$
Kevin C.

ah bugger. Thinking I should have had more sleep. (Sleepy)
Thanks