Question

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

Answer 1

`~~ -0.01`

Explanation:

We know that the slope of the tangent at `x = 2` is `f'(2)`. Now

`f'(x) = 2 e^{2 x} -3 x^2`

so that

`f'(2) = 2 e^4-12~~97.2`

Thus the slope of the line normal to this is `-1/97.2 ~~ -0.01`