Question

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

Answer 1

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'(x) = e^(-x^3-x)+(x-1)(-3x^2-1)e^(-x^3-x)` by the product rule.

Therefore the gradient at `x=1` is given by:
`f'(1)=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(1)=0`

Finally,
In general a straight line is given by:
`y-y_1=m(x-x_1)`
By substituting what we know:
`y-0=-e^2(x-1)`
Finally:
`y=-e^2x+e^2`