What is the equation of the line normal to #f(x)=xe^x# at #x=7#?

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Answer 1 · 2016-02-02T14:59:45

Equation of normal line

#y = (7(8e^14 + 1) - x)/(8e^{7})#

#f(7) = 7e^7#

The normal line has to pass through the point #(7,f(7))#.

#f'(x) = (x+1)e^x#

#f'(7) = 8e^7#

Since the normal line is perpendicular to the tangent line, we have the gradient of the normal line, #m#, given by

#m = frac{-1}{f'(7)} = -frac{1}{8}e^{-7}#.

The #y#-intercept of the line is given by

#c = y - mx#

#= 7e^7 - (-frac{1}{8}e^{-7})(7)#

#= 7e^{-7}(e^14 + 1/8)#

Equation of normal line in slope-intercept form

#y = -frac{1}{8}e^{-7}x + 7e^{-7}(e^14 + 1/8)#