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

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Answer 1 · 2017-01-21T04:51:56

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

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

#f=1.265, at x = 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]}