Question

What are the local extrema, if any, of `f (x) =x^3-3x+6`?

Answer 1

`x^3-3x+6` has local extrema at `x=-1` and `x=1`

Explanation:

The local extrema of a function occur at points where the first derivative of the function is `0` and the sign of the first derivative changes.
That is, for `x` where `f'(x) = 0` and either `f'(x-varepsilon) <= 0 and f'(x+varepsilon) >= 0` (local minimum) or
`f'(x-varepsilon) >= 0 and f'(x+varepsilon) <= 0` (local maximum)

To find the local extrema, then, we need to find the points where `f'(x) = 0`.

`f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x+1)(x-1)`
so
`f'(x) = 0 <=> 3(x+1)(x-1) = 0 <=> x=+-1`

Looking at the sign of `f'` we get
`{(f'(x) > 0 if x < -1), (f'(x) < 0 if -1 < x < 1), (f'(x) > 0 if x > 1):}`

So the sign of `f'` changes at each of `x = -1` and `x = 1` meaning there is a local extremum at both points.

Note: From the change in signs, we can further tell that there is a local maximum at `x = -1` and a local minimum at `x = 1`.