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

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

#### Explanation:

Let's find the equation of the slope of the function.

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

So, Slope of the Function $f \left(x\right)$ at $x = 1$ is

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

We know, The condition of perpendicularity of two straight lines in 2D Plane is,

${m}_{1} {m}_{2} = - 1$, where ${m}_{1}$ and ${m}_{2}$ are the slopes of the straight lines respectively.

Let the Required Normal Line be $y = m x + c$.

According to the question,

$- 2 m = 1 \Rightarrow m = - \frac{1}{2}$ [Condition of Perpendicularity]

So, The equation is now $y = - \frac{1}{2} x + c$...................(i)

Now, $f \left(x\right)$ at $x = 1$ :- $f \left(1\right) = \frac{1}{{1}^{2} - 2} = - 1$

Let's put $y = - 1$ and $x = 1$ in eq(i).

So, We get,

$\textcolor{w h i t e}{\times x} - 1 = - \frac{1}{2} \left(1\right) + c$

$\Rightarrow - 1 = - \frac{1}{2} + c$

$\Rightarrow c = \frac{1}{2} - 1$

$\Rightarrow c = - \frac{1}{2}$

So, The final equation is :-

$\textcolor{w h i t e}{\times x} y = - \frac{1}{2} x - \frac{1}{2}$

$\Rightarrow y = \frac{- x - 2}{2}$

$\Rightarrow y = - \frac{x + 2}{2}$

Hope this helps.