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

Answer 1

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#

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#.

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Answer 2

Emilia Aguilar

The equation of the line normal to f(x) = ln(x^2) - 1/x^2 at x = -2 is y = -4x - 3.

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Answer from HIX Tutor

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