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

1 Answer
Jan 4, 2016

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

Explanation:

The normal line to the graph of #f# at the point #(a,f(a))# is perpendicular ("normal") to the tangent line at that point. The tangent line has slope #f'(a)#, so the normal line has slope #-1/(f'(a))# (the "negative reciprocal" of #f'(a)#).

Hence, the equation of the normal line to #f# at the point #(a,f(a))# has equation

#y=f(a)-1/(f'(a))(x-a)#.

In the present case, with #f(x)=1/(2x^2-1)#, the quotient rule leads to #f'(x)=-(4x)/((2x^{2}-1)^2)#. Hence, #f(1)=1/(2-1)=1# and #f'(1)=-(4)/((2-1)^2)=-4#, making #-1/(f'(a))=1/4#.

This means the answer is #y=1+1/4(x-1)=1/4 x+3/4#.