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

Jan 21, 2017

$16.58 x + y - 34.43 = 0$. See the normal-inclusive graph.

#### Explanation:

$f \sqrt{x + \frac{1}{x}} = x$.

$f = 1.265 , a t x = 2$. And,

$f ' \sqrt{x + \frac{1}{x}} + \frac{1}{2} f \frac{1 - \frac{1}{x} ^ 2}{\sqrt{x + \frac{1}{x}}} = 1.$ that gives

$f ' = 0.060 , a t x = 2.$

The slope of the normal is -1 / f'=--16.58, nearly.

So, the equation to the normal at P(2, 1.265) is

$y - 1.265 = - 16.58 \left(x - 2\right)$, that gives

$16.58 x + y - 34.43 = 0$

graph{(y-x/sqrt(x+1/x))(y+16.6x-34.4)((x-2)^2+(y-1.3)^2-.01)=0 [-10, 10, -5, 5]}