Question

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

Answer 1

Slope of the normal to `f(x)` at `x=0` is `1/2`

Explanation:

`f(x) = e^(-x)+x^2-x`

Differentiating:
`f'(x) =-e^(-x)+2x-1`

To find the slope of `f(x)` at `x=0`:
`f'(0) = -1+0-1 =-2`

Hence slope of the tangent `(m)` to `f(x)` at `x=0` is `-2`

Since the slope of the normal at the same point = `-1/m`
Therefore the slope of the normal to `f(x)` at `x=0 = -1/(-2) = 1/2`