Question

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

Answer 1

`e^2/(2-e^2)`

Explanation:

Differentiating,

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

`f'(0)=(-2)e^(-2)+1=1-2/e^2=(e^2-2)/e^2`

Slope of the tangent at `x=0` is `(e^2-2)/e^2`

Slope of the normal at `x=0` is `e^2/(2-e^2)`