What is the equation of the line normal to # f(x)=x^2/e^x-x/e^(x^2) # at # x=0#?

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Answer 1 · 2015-12-29T07:04:41

#y = x#

#color(blue)("Graph of "f(x))#
graph{(x^2)/(e^x)-x/(e^(x^2)}

#f(0) = 0#

Since the Normal passes through #(0,0)#, it has the form of #y = mx#.

Since the Normal is perpendicular to the tangent of #f# at #x = 0#, #m*f'(0) = -1#.

To find #f'(0)#, we first find the general #f'(x)#, and then substitute #x = 0#.

#f'(x) = frac{d}{dx}( frac{x^2}{e^x} - frac{x}{ e^{x^2} } )#

#= frac{e^x(2x) - x^2(e^x)}{(e^x)^2} - frac{e^{x^2}(1) - x(2xe^{x^2})}{(e^{x^2})^2}#

#= frac{2x - x^2}{e^x} - frac{1 - 2x^2}{e^{x^2}}#

#f'(0) = -1#

Therefore, #m = frac{-1}{-1} = 1#

Equation of Normal: #y = x#

#color(blue)("Graph of Normal")#
graph{x}