Question

What are the points of inflection, if any, of `f(x) =x^3 - 3x^2 + 3x`?

Answer 1

`f(x)=x^3-3x^2+3x` has a single point of inflection at `x=1`

Explanation:

At a point of inflection the slope of the function must be equal to zero.
(Note that a slope of zero does not necessarily indicate a point of inflection; it could be a local minimum or local maximum).

If `f'(x)=3x^2-6x+3=0`
then
`color(white)("XXX")x^2-2x+1=0`

`color(white)("XXX")(x-1)^2=0`

so the only candidate as a possible point of inflection is `x=1`

There are numerous ways to check if this point is a local minimum, point of inflection, or local maximum.
First and second derivative tests are often suggested.
Here is an alternative:

Suppose `barx` is our candidate to be tested
Pick any value `v_l < barx` such that `v_l` is greater than any other value `x_l < barx` for which `f'(x_l)=0`
and
any value `v_r > barx` such that `v_r` is less than any other value `x_r > barx` for which `f'(x_r)=0`
`color(white)("XXX")`(this is simpler than it first sounds).

For our example, since there are no values other than `x=1` for which `f'(x)=0` we can pick any values `v_l < 1` and `v_r > 1`.

#{: ((f(barx) > f(v_l))color(white)("X") & color(white)("X")(f(barx) > f(v_r)),rArr,f(barx)" is a maximum"), (,,), ((f(barx) < f(v_l))color(white)("X") & color(white)("X")(f(barx) < f(v_r)),rArr,f(barx)" is a minimum"), (,,), ("otherwise",rArr,f(barx)" is a point of inflection") :}#

For this example, choosing `v_l=0` and `v_r=2` (with candidate `barx=1`)

`color(white)("XXX")f(0)=(0)^3-3(0)^2+3(0)=0`
`color(white)("XXX")f(1)=(1)^3-3(1)^2+3(1)= 1`
`color(white)("XXX")f(2)=2^3-3(2)^2+3(2) = 8-12+6 = 2`

`f(barx=1) > f(v_l=0)`
but
`f(barx=1) < f(v_r=2)`

Therefore `barx=1` is a point of inflection.