# What is the slope of the line normal to the tangent line of f(x) = 1/(x^2-2x+4)  at  x= 3 ?

Mar 12, 2016

slope of the Normal to the tangent at x=3 is $- \frac{1}{f} ^ ' \left(3\right) = \frac{49}{4}$

#### Explanation:

First find ${f}^{'} \left(x\right)$ and then find its value putting x= 3

${f}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} {\left({x}^{2} - 2 x + 4\right)}^{-} 1$
$= - {\left({x}^{2} - 2 x + 4\right)}^{-} 2 \frac{d}{\mathrm{dx}} \left({x}^{2} - 2 x + 4\right)$
=-(x^2-2x+4)^-2(d/(dx)(x^2)-2d/(dx)(x)+d/(dx)(4)))
$= - {\left({x}^{2} - 2 x + 4\right)}^{-} 2 \left(2 x - 2\right)$
$\therefore {f}^{'} \left(x\right) = - {\left({x}^{2} - 2 x + 4\right)}^{-} 2 \left(2 x - 2\right)$
$\therefore {f}^{'} \left(3\right) = - {\left({3}^{2} - 2 \cdot 3 + 4\right)}^{-} 2 \left(2 \cdot 3 - 2\right) = - \frac{4}{49}$

Hence slope of the tangent at x=3 is ${f}^{'} \left(3\right) = - \frac{4}{49}$

slope of the Normal to the tangent at x=3 is $- \frac{1}{f} ^ ' \left(3\right) = \frac{49}{4}$