Question

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

Answer 1

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

Explanation:

To begin with , for x=-1, f(x) would be `-1-e^(1+1) -> -1-e^2`. So the normal has to be found at point `(-1, -1-e^2)`

Now slope of f(x) would be `f'(x)=-1/x^2 -e^(x^2-x^3) (2x-3x^2)`. The slope at x=-1 would be `f'(-1)= -1-e^2 (-2-3) = -1+5e^2`

The slope of the normal line would thus be `1/(1-5e^2)`

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

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