Question

What is the equation of the normal line of `f(x)= 1/xe^(-x^3+x^2)` at `x=-1`?

Answer 1

`5e^2y+5e^4+x+1=0`

Explanation:

You start by finding the first derivative .
`y'=(-3x+2)e^(-x^3+x^2)`

by substituting with `x=-1` in both the function and the first derivative you get the y co_ordinate of the point that lies on the normal line and the slope of the tangent to the curve ( m )

`y=1/-1e^(2)`
so the y co_ordinate of the point =`-e^2`

and the slope of the tangent ( m ) at `x=-1` equals:
`y'=5e^2`
but We don't want the slope of the tangent, We want the slope of the normal.

The slope of the normal =`-1/m`

so it will be `-1/5e^-2`

and to get the equation of the straight line

`y-(-e^2)=-1/5e^-2(x-(-1)`

and by simplification you get:
`5e^2y+5e^4+x+1=0`