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

Feb 6, 2016

$f \left(x\right) = {x}^{3} - 3 {x}^{2} + 3 x$ 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 $f ' \left(x\right) = 3 {x}^{2} - 6 x + 3 = 0$
then
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 2 x + 1 = 0$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} = 0$

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 $\overline{x}$ is our candidate to be tested
Pick any value ${v}_{l} < \overline{x}$ such that ${v}_{l}$ is greater than any other value ${x}_{l} < \overline{x}$ for which $f ' \left({x}_{l}\right) = 0$
and
any value ${v}_{r} > \overline{x}$ such that ${v}_{r}$ is less than any other value ${x}_{r} > \overline{x}$ for which $f ' \left({x}_{r}\right) = 0$
$\textcolor{w h i t e}{\text{XXX}}$(this is simpler than it first sounds).

For our example, since there are no values other than $x = 1$ for which $f ' \left(x\right) = 0$ 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 $\overline{x} = 1$)

$\textcolor{w h i t e}{\text{XXX}} f \left(0\right) = {\left(0\right)}^{3} - 3 {\left(0\right)}^{2} + 3 \left(0\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} f \left(1\right) = {\left(1\right)}^{3} - 3 {\left(1\right)}^{2} + 3 \left(1\right) = 1$
$\textcolor{w h i t e}{\text{XXX}} f \left(2\right) = {2}^{3} - 3 {\left(2\right)}^{2} + 3 \left(2\right) = 8 - 12 + 6 = 2$

$f \left(\overline{x} = 1\right) > f \left({v}_{l} = 0\right)$
but
$f \left(\overline{x} = 1\right) < f \left({v}_{r} = 2\right)$

Therefore $\overline{x} = 1$ is a point of inflection.