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

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Answer 1 · 2016-03-31T05:58:10

Equation of normal is #x-y-1=0#

At #x=1#, #f(x)=(x-1)/(x^2-2)=(1-1)/(1^2-2)=0#

So normal or tangent will be at #x=1# will be at #(1,0)# on the curve.

Slope of tangent is given by #f'(x)# which is given by

#f'(x)=((x^2-2)d/dx(x-1)-(x-1)d/dx(x^2-2))/(x^2-2)^2# or

= #((x^2-2)xx1-(x-1)xx2x)/(x^2-2)^2=(x^2-2-2x^2+2x)/(x^2-2)^2# or

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

Hence, at #x=1#, #f'(x)=-(1^2-2*1+2)/(1-2)^2=-1#

As slope of tangent is #-1#, slope of normal would be #-1/-1=1#

Hence using point slope form equation of normal is

#(y-0)=1(x-1)# or #x-y-1=0#