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

Nov 22, 2017

$y = \frac{1}{3} x - \frac{4}{3}$

#### Explanation:

In order to find the normal to the tangent we first need to find the gradient of the tangent line at $x = 1$

Differentiate $f \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(x {\left({x}^{2} - 2\right)}^{- 1}\right) = {\left({x}^{2} - 2\right)}^{- 1} - {\left({x}^{2} - 2\right)}^{- 2} \cdot 2 x \cdot x$

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

Gradient of tangent at $x = 1$

$- \frac{{1}^{2} + 2}{{1}^{2} - 2} ^ 2 = - 3$

$- 3 \cdot {m}_{2} = - 1 \implies {m}_{2} = \frac{1}{3}$

Coordinates at $x = 1$

$\left(1 , - 1\right)$

Equation of normal:

$y + 1 = \frac{1}{3} \left(x - 1\right) \implies y = \frac{1}{3} x - \frac{4}{3}$

Graph: