Question

How do you find the slope of the tangent line of `f(x) = 3-2x^3` at (-1, 5)?

Answer 1

I found: slope`=-6`

Explanation:

The slope of the tangent will be the derivative of your function evaluated at `x=-1` (the `x` coordinate of the tangence point).
So:
`f'(x)=0-6x^2=-6x^2`
evaluated at `x=-1`:
`f'(-1)=-6`
So the slope of the tangent will be `-6`.