What is the equation of the tangent line of #f(x) = e^(x)/(x^2-x# at #x=3#?

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Answer 1 · 2018-05-27T09:44:19

#y=e^3/36x+e^3/12#

#f(x)=e^x/(x^2-x)#

#D_f={AAx##in##RR##:x^2-x!=0}=(-oo,0)uu(0,1)uu(1,+oo)=RR-{0,1}#

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

#(e^x(x^2-x)-e^x(2x-1))/(x^2-x)^2=(x^2e^x-xe^x-2xe^x+e^x)/(x^2-x)^2=#

#(x^2e^x-3xe^x+e^x)/(x^2-x)^2#

For the equation of the tangent line at #A(3,f(3))# we require the values

#f(3)=e^3/6#

#f'(3)=(9e^3-9e^3+e^3)/36=e^3/36#

The equation will be

#y-f(3)=f'(3)(x-3)# #<=>#

#y-e^3/6=e^3/36(x-3)# #<=>#

#y-e^3/6=e^3/36x-cancel(3)e^3/cancel(36)# #<=>#

#y=e^3/36x-e^3/12+e^3/6# #<=>#

#y=e^3/36x+e^3/12#

and a graph

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