# What is the slope of the line normal to the tangent line of f(x) = 1/(x^2-x+1)  at  x= 2 ?

##### 1 Answer
Jan 8, 2016

slope of normal = 3

#### Explanation:

start by rewriting the function as

$f \left(x\right) = {\left({x}^{2} - x + 1\right)}^{-} 1$

now using the chain rule to differentiate :

$f ' \left(x\right) = - 1 {\left({x}^{2} - x + 1\right)}^{-} 2. \frac{d}{\mathrm{dx}} \left({x}^{2} - x + 1\right)$

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

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

now f'(x) = slope of tangent line and evaluating gives :

$f ' \left(2\right) = \frac{- \left(4 - 1\right)}{4 - 2 + 1} ^ 2 = - \frac{3}{9} = - \frac{1}{3}$

using ${m}_{1} {m}_{2} = - 1$ for 2 lines which are perpendicular to each other.

then slope of the normal to the tangent is m = 3