Question

Question #cbfb8

Answer 1

There is one critical number, at `x= 1`, which is neither a maximum nor a minimum.

Explanation:

Start by finding the derivative by the power rule.

`f'(x) = 3x^2 - 6x + 3`

The critical points will occur when the derivative equals `0` or is undefined. This is a quadratic function, so it is defined on all of `x`.

`0 = 3x^2 - 6x + 3`

`0 = 3(x^2 - 2x + 1)`

`0 = (x -1)(x - 1)`

`x = 1`

Next, we must verify whether this is a local minimum or a local maximum. Note that this function will never have an absolute maximum/minimum, as shows the table below.

We can verify whether this point is a local min or a local max by finding the intervals of increase/decrease.

Test point 1: `x = 0`

`f'(0) = 3(0)^2 - 6(0) + 3 = 3`

Test point 2: `x = 3`

`f'(3) = 3(3)^2 - 6(3) + 3 = 27 - 18 + 3 = 12`

Since `f'(x) > 0` on both sides of `x = 1`, our single critical point is neither a maximum nor a minimum.

Hopefully this helps!