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

1 Answer
Nov 22, 2017

#y=1/3x-4/3#

Explanation:

In order to find the normal to the tangent we first need to find the gradient of the tangent line at #x=1#

Differentiate #f(x)#

#d/dx(x(x^2-2)^(-1))=(x^2-2)^(-1)-(x^2-2)^(-2)*2x*x#

#->=1/((x^2-2))- (2x^2)/(x^2-2)^2=((x^2-2)-2x^2)/(x^2-2)^2=(-x^2-2)/(x^2-2)^2=- (x^2+2)/(x^2-2)^2#

Gradient of tangent at #x=1#

#- (1^2+2)/(1^2-2)^2=-3#

gradient of normal:

#-3*m_2=-1=>m_2=1/3#

Coordinates at #x=1#

#( 1 , -1 )#

Equation of normal:

#y+1=1/3(x-1)=>y=1/3x-4/3#

Graph:

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