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

Apr 6, 2018

$5 {e}^{2} y + 5 {e}^{4} + x + 1 = 0$

#### Explanation:

You start by finding the first derivative .
$y ' = \left(- 3 x + 2\right) {e}^{- {x}^{3} + {x}^{2}}$

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 )

$y = \frac{1}{-} 1 {e}^{2}$
so the y co_ordinate of the point =$- {e}^{2}$

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

The slope of the normal =$- \frac{1}{m}$

so it will be $- \frac{1}{5} {e}^{-} 2$

and to get the equation of the straight line

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

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