Question

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

Answer 1

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]}