What is the slope of the line normal to the tangent line of #f(x) = xe^-x+x^2-x # at # x= 0 #?

1 Answer
Apr 27, 2018

Normal is vertical and hence its slope is not defined.

Explanation:

Slope of a tangent to the curve #f(x)# is given by the value of #f'(x)# at that point.

Here #f(x)=xe^(-x)+x^2-x# and hence

#f'(x)=1*e^(-x)+x*(-e^(-x))+2x-1#

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

and at #x=0# we have #f'(0)=e^0(1-0)+2*0-1=1+0-1=0#

i.e. at #x=0#, tangent is horizontal

and hence normal, which is always perpendicular to tangent

and hence its slope is not defined.

graph{xe^(-x)+x^2-x [-2.5, 2.5, -1.25, 1.25]}