# How do you find the tangent line approximation to f(x)=1/x near x=1 ?

Aug 14, 2014

$L \left(x\right) = 2 - x$

We find the answer from the linear approximation $L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$. For linear approximations, we want both $f \left(a\right)$ and $f ' \left(a\right)$ so that they are easy to calculate. In this case, $f \left(1\right)$ and $f ' \left(1\right)$ are easy.

To find the derivative of

$f \left(x\right) = \frac{1}{x}$

it is best to rewrite it in power form:

$f \left(x\right) = {x}^{- 1}$

Using the power rule for derivatives:

$f ' \left(x\right) = - {x}^{- 2}$

So substituting, $a = 1$:

$f \left(a\right) = \frac{1}{1}$

$f ' \left(1\right) = - \frac{1}{1} ^ 2$

Substituting this into $L \left(x\right)$,

$L \left(x\right) = 1 - 1 \left(x - 1\right)$

$= 1 - x + 1$

$= 2 - x$