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

Dec 13, 2016

$y - \left(- 1 - {e}^{2}\right) = \frac{1}{1 - 5 {e}^{2}} \left(x + 1\right)$

#### Explanation:

To begin with , for x=-1, f(x) would be $- 1 - {e}^{1 + 1} \to - 1 - {e}^{2}$. So the normal has to be found at point $\left(- 1 , - 1 - {e}^{2}\right)$

Now slope of f(x) would be $f ' \left(x\right) = - \frac{1}{x} ^ 2 - {e}^{{x}^{2} - {x}^{3}} \left(2 x - 3 {x}^{2}\right)$. The slope at x=-1 would be $f ' \left(- 1\right) = - 1 - {e}^{2} \left(- 2 - 3\right) = - 1 + 5 {e}^{2}$

The slope of the normal line would thus be $\frac{1}{1 - 5 {e}^{2}}$

The equation(point-slope form) of the normal line would thus be

$y - \left(- 1 - {e}^{2}\right) = \frac{1}{1 - 5 {e}^{2}} \left(x + 1\right)$