What is the equation of the line normal to # f(x)=lnx-1/x# at # x=3#?

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Answer 1 · 2016-01-08T20:59:06

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

now the line equation is:

#(y-y_0)=m(x-x_0)#

where:

#x_0=3#
#y_0=f(x_0)#
#m=-1/(f'(x_0))# because we looking for the normal line

then:

#f'(3)=(3-1)/9=2/9#
#-1/(f'(x))=-9/2#
#f(x_0)=ln3-1/3#

#y-(ln3-1/3)=-9/2(x-3)#
#y=-9/2x-27/2+(ln3-1/3)#