# If 1/3(x-1)^3+2, what are the points of inflection, concavity and critical points?

Nov 8, 2017

The point of inflection is $= \left(1 , 2\right)$. The interval of concavity is $x \in \left(- \infty , 1\right)$. The interval of convexity is $\left(1 , + \infty\right)$

#### Explanation:

Let $f \left(x\right) = \frac{1}{3} {\left(x - 1\right)}^{3} + 2$

Let's develop $f \left(x\right)$

$f \left(x\right) = \frac{1}{3} \left({x}^{3} - 3 {x}^{2} + 3 x - 1\right) + 2$

$= {x}^{3} / 3 - {x}^{2} + x + \frac{5}{3}$

Calculate the first derivative

$f ' \left(x\right) = {x}^{2} - 2 x + 1 = {\left(x - 1\right)}^{2}$

The critical points are when $f ' \left(x\right) = 0$

That is,

${\left(x - 1\right)}^{2} = 0$, $\implies$, $x = 1$

Now, calculate the second derivative

$f ' ' \left(x\right) = 2 x - 2$

The points of inflections are when $f ' ' \left(x\right) = 0$

That is,

$2 x - 2 = 0$, $\implies$, $x = 1$

Make a sign chart

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , 1\right)$$\textcolor{w h i t e}{a a a a}$$\left(1 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$S i g n f ' ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

graph{1/3(x-1)^3+2 [-8.335, 9.445, -2.355, 6.534]}