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

Apr 27, 2018

Normal is vertical and hence its slope is not defined.

Explanation:

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

Here $f \left(x\right) = x {e}^{- x} + {x}^{2} - x$ and hence

$f ' \left(x\right) = 1 \cdot {e}^{- x} + x \cdot \left(- {e}^{- x}\right) + 2 x - 1$

= ${e}^{- x} \left(1 - x\right) + 2 x - 1$

and at $x = 0$ we have $f ' \left(0\right) = {e}^{0} \left(1 - 0\right) + 2 \cdot 0 - 1 = 1 + 0 - 1 = 0$

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]}