Question

How do you locate the critical point, and determine the maxima and minima of `y=x^3-6x^2+9x-8`?

Answer 1

`f(x)=y=x^3-6x^2+9x-8`

Critical points:
`c` is a critcal point for `f` if `c` is in the domain of `f` and either `f'(c)=0` or `f'(c)` does not exist.

For this function, the domain is `(-oo, oo)` and `f'(x)=3x^2-12x+9` exists for all real `x`.
So the critical points for this `f` will be the zeros of `f'`.

`f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-3)(x-1)=0`
at `x=1`, `x=3`
These are the critical points for this `f`.

Testing the critical points for extrema

You may test critical points using either the first or the second derivative test for local extrema. (A.k.a relative extrema).

Using the first derivative test (for local extrema) we look at the intervals defined by the critical points and test for `f'<0 " or" >0` on each interval.
You can choose any test numbers you like, I'll choose here.

`f'(x)=3x^2-12x+9=3(x-3)(x-1)=0`

Intervals Test numbers and sign of `f'`

`(-oo, 1)`; test `x=0`; sign `f'(0)=3(0-3)(0-1)=+" number" >0`
`(1, 3)`; test `x=2`; sign `f'(2)=3(2-3)(2-1)=-" number" <0`
`(3, oo)`; test `x=5`; sign `f'(5)=3(5-3)(5-1)=+" number" >0`

`f'` changes from + to - as we pass `x=1`, so `f(1)=(1)^3-6(1)^2+9(1)-8=-4` is a local maximum.

`f'` changes from - to + as we pass `x=3`, so `f(3)=(3)^3-6(3)^2+9(3)-8=-8` is a local maximum.

Arithmetic:
`f(3)=(3)^3-6(3)^2+9(3)-8=(3)^3-2(3)^3+3^3-8=(0)(3)^3-8=-8`

It's not needed to answer the question, but here is the graph:

graph{y=x^3-6x^2+9x-8 [-10.45, 21.6, -14.22, 1.8]}