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

Jan 4, 2016

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

#### Explanation:

The normal line to the graph of $f$ at the point $\left(a , f \left(a\right)\right)$ is perpendicular ("normal") to the tangent line at that point. The tangent line has slope $f ' \left(a\right)$, so the normal line has slope $- \frac{1}{f ' \left(a\right)}$ (the "negative reciprocal" of $f ' \left(a\right)$).

Hence, the equation of the normal line to $f$ at the point $\left(a , f \left(a\right)\right)$ has equation

$y = f \left(a\right) - \frac{1}{f ' \left(a\right)} \left(x - a\right)$.

In the present case, with $f \left(x\right) = \frac{1}{2 {x}^{2} - 1}$, the quotient rule leads to $f ' \left(x\right) = - \frac{4 x}{{\left(2 {x}^{2} - 1\right)}^{2}}$. Hence, $f \left(1\right) = \frac{1}{2 - 1} = 1$ and $f ' \left(1\right) = - \frac{4}{{\left(2 - 1\right)}^{2}} = - 4$, making $- \frac{1}{f ' \left(a\right)} = \frac{1}{4}$.

This means the answer is $y = 1 + \frac{1}{4} \left(x - 1\right) = \frac{1}{4} x + \frac{3}{4}$.