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

May 20, 2016

See below.

Explanation:

I'll first explain my reasoning for what I'm about to do:
Find the derivative of the function
Find the gradient of the tangent at $x = 1$
Find the gradient of the normal line at $x = 1$
Find the y-coordinate at $x = 1$
Find the equation of the normal line with the above information

Firstly $f ' \left(x\right) = {e}^{- {x}^{3} - x} + \left(x - 1\right) \left(- 3 {x}^{2} - 1\right) {e}^{- {x}^{3} - x}$ by the product rule.

Therefore the gradient at $x = 1$ is given by:
$f ' \left(1\right) = {e}^{-} 2 + 0$

Therefore the gradient of the normal line at $x = 1$ is:
$- {e}^{2}$

The y-coordinate of the function at $x = 1$ is given by:
$f \left(1\right) = 0$

Finally,
In general a straight line is given by:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
By substituting what we know:
$y - 0 = - {e}^{2} \left(x - 1\right)$
Finally:
$y = - {e}^{2} x + {e}^{2}$