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

Mar 31, 2016

Equation of normal is $x - y - 1 = 0$

#### Explanation:

At $x = 1$, $f \left(x\right) = \frac{x - 1}{{x}^{2} - 2} = \frac{1 - 1}{{1}^{2} - 2} = 0$

So normal or tangent will be at $x = 1$ will be at $\left(1 , 0\right)$ on the curve.

Slope of tangent is given by $f ' \left(x\right)$ which is given by

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

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

= $- \frac{{x}^{2} - 2 x + 2}{{x}^{2} - 2} ^ 2$

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

As slope of tangent is $- 1$, slope of normal would be $- \frac{1}{-} 1 = 1$

Hence using point slope form equation of normal is

$\left(y - 0\right) = 1 \left(x - 1\right)$ or $x - y - 1 = 0$