# If  f(x) = x^3 - 6x^2 + 9x +1, what are the points of inflection, concavity and critical points?

Feb 18, 2018

Here, the critical points are
$\left(1 , 5\right) , \text{where the slope is zero}$
$\text{ and curvature is negative, thus being a maximum}$$\text{ representing concave down}$

$\left(3 , 1\right) , \text{where the slope is zero}$
$\text{ and curvature is positive, thus being a minimum }$$\text{representing concave up}$

However, the point
$\left(2 , 3\right) , \text{where the curvature is zero}$
$\text{ and curve is changing from concave down to concave up}$$\text{known as point of inflection}$

#### Explanation:

$\text{Given:}$

$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 9 x + 1$

$f ' \left(x\right) = 3 {x}^{2} - 12 x + 9$

$\text{Solving for x where f'(x)=0,}$

$3 {x}^{2} - 12 x + 9 = 0$

$\text{Dividing by 3}$

${x}^{2} - 4 x + 3 = 0$

${x}^{2} - 3 x - 1 x + 3 = 0$

$x \left(x - 3\right) - 1 \left(x - 3\right) = 0$

$\left(x - 1\right) \left(x - 3\right) = 0$

$x = 1 , x = 3. \text{form the points where the slope is zero}$

$f ' \left(x\right) = 3 {x}^{2} - 12 x + 9$

$f ' ' \left(x\right) = 6 x - 12$

$\text{Solving for x where f''(x) is zero}$

$6 x - 12 = 0$

$6 \left(x - 2\right) = 0$

$x - 2 = 0$

$x = 2. \text{form the point where the curvature is zero.}$
$\text{The point where the curvature changes its sign}$

$x = 2 , \text{forms a point of inflexion}$

$x = 1 , \text{the curvature is} ,$
$6 x - 12 = 6 \times 1 - 12 = 6 - 12 = - 6$

$x = 3 , \text{the curvature is} ,$
$6 x - 12 = 6 \times 3 - 12 = 18 - 12 = + 6$

$\text{From x=1, to x=2, the curvature is negative}$
$\text{ indicating concave down}$

$\text{From x=2, to x=3, the curvature is positive}$
$\text{ indicating concave up}$

The function takes the values as evaluated below

$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 9 x + 1$

$f \left(1\right) = {\left(1\right)}^{3} - 6 {\left(1\right)}^{2} + 9 \left(1\right) + 1 = 1 - 6 + 9 + 1 = - 5 + 10 = 5$

$f \left(2\right) = {\left(2\right)}^{3} - 6 {\left(2\right)}^{2} + 9 \left(2\right) + 1 = 8 - 24 + 18 + 1 = - 16 + 19 = 3$

$f \left(3\right) = {\left(3\right)}^{3} - 6 {\left(3\right)}^{2} + 9 \left(3\right) + 1 = 27 - 54 + 27 + 1 = - 27 + 28 = 1$

Here, the critical points are
$\left(1 , 5\right) , \text{where the slope is zero}$
$\text{ and curvature is negative, thus being a maximum}$$\text{ representing concave down}$

$\left(3 , 1\right) , \text{where the slope is zero}$
$\text{ and curvature is positive, thus being a minimum }$$\text{representing concave up}$

However, the point
$\left(2 , 3\right) , \text{where the curvature is zero}$
$\text{ and curve is changing from concave down to concave up}$