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

1 Answer
Feb 12, 2017

The normal line to this function at the given point has equation y=4/5 (x+2)+ln(4)-1/4=4/5 x+ln(4)+27/20

Explanation:

First note that f(-2)=ln(4)-1/4, so the graph of the normal line goes through the point (-2,ln(4)-1/4).

Second, f'(x)=2/x+2/x^3 so that f'(-2)=-1-1/4=-5/4 is the slope of the tangent line to the graph of f at the given point.

Since the normal line has slope equal to the negative reciprocal of the tangent line (since they are perpendicular), the normal line has slope -1/(-5/4)=4/5.

Putting all this together gives the equation of the normal line to be y=4/5 (x+2)+ln(4)-1/4=4/5 x+ln(4)+27/20.