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

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Answer 1 · 2017-01-08T13:59:02

#y = 8/e^2 -e^2/4(x-2)#

The equation of the line normal to the graph of the function:

#y=f(x)#

at ath point #(bar x, f(barx)) # is given by:

#y=f(bar x)-1/(f'(barx)) (x-barx)#

In our case:

#f(x) = x^3/e^x=x^3e^(-x)#

#f(2) = 8/e^2#

#f'(x) = 3x^2e^-x -x^3e^-x#

#f'(2) = 12/e^2-8/e^2=4/e^2#

So the normal line is:

#y = 8/e^2 -e^2/4(x-2)#