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

1 Answer
Jan 21, 2017

16.58 x+y-34.43=016.58x+y34.43=0. See the normal-inclusive graph.

Explanation:

fsqrt(x+1/x)=xfx+1x=x.

f=1.265, at x = 2f=1.265,atx=2. And,

f'sqrt(x+1/x)+1/2f(1-1/x^2)/(sqrt(x+1/x))=1. that gives

f'=0.060, at 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(x-2), that gives

16.58x+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]}