What is the equation of the normal line of #f(x)=xe^x# at #x=3 #?

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Answer 1 · 2018-07-19T15:15:57

#y-3e^3=-1/{4e^3}(x-3)#

Substituting #x=3# in the given function, we get y-coordinate of point is obtained as follows

#y=f(3)=3e^3#

Differentiating given function w.r.t. #x#, the slope of tangent #dy/dx# is

#dy/dx=f'(x)#

#=d/dx(xe^x)#

#=xe^x+e^x#

hence the slope of tangent at #x=3# is

#=f'(3)#

#=3e^3+e^3#

#=4e^3#

hence the slope #m# of normal at the same point #x=3# is given as

#m=-1/{f'(3)}#

#=-1/{4e^3}#

Now, the equation of normal at the point #(3, 3e^3)# & having slope #m=-1/{4e^3}# is given as follows

#y-3e^3=-1/{4e^3}(x-3)#