# How do you find all critical point and determine the min, max and inflection given f(x)=x^3+x^2-x?

Nov 14, 2016

Critical Points are:
$\left(- 1 , 1\right)$ min
$\left(\frac{1}{3} , - \frac{5}{27}\right)$ max

#### Explanation:

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

To identify the critical vales, we differentiate and find find values of $x$ st $f ' \left(x\right) = 0$

 { ( f'(x) < 0, => f(x) " is decreasing" ), ( f'(x) = 0, => f(x) " is stationary" ), ( f'(x) > 0, => f(x) " is increasing" ) :}

Differentiating wrt $x$' we have:

$f ' \left(x\right) = 3 {x}^{2} + 2 x - 1$ .... [1]

At a critical point, $f ' \left(x\right) = 0$

$f ' \left(x\right) = 0 \implies 3 {x}^{2} + 2 x - 1 = 0$
$\therefore \left(3 x - 1\right) \left(x + 1\right) = 0$
$x = - 1 , \frac{1}{3}$

Ton find the y-coordinate we substitute the required value into $f \left(x\right)$
$x = - 1 \implies f \left(- 1\right) = - 1 + 1 - \left(- 1\right) = 1$
$x = \frac{1}{3} \implies f \left(\frac{2}{3}\right) = \frac{1}{27} + \frac{1}{9} + \frac{1}{3} = - \frac{5}{27}$

So the critical points are $\left(- 1 , 1\right)$ and $\left(\frac{1}{3} , - \frac{5}{27}\right)$

We identify the nature of these critical points by looking at the sign of second derivative, and

 { ( f''(x) < 0, => f'(x) " is decreasing" => "maximum" ), ( f''(x) = 0, => f'(x) " is stationary" => "inflection" ), ( f''(x) > 0, => f'(x) " is increasing" => "minimum" ) :}

Differentiating [1] wrt $x$ gives s;

$f ' ' \left(x\right) = 6 x + 2$
$x = - 1 \implies f ' ' \left(- 2\right) = - 6 + 2 < 0$, ie a maximum
$x = \frac{1}{3} \implies f ' ' \left(1\right) = 2 + 2 > 0$, ie a minimum

Incidental, As this is a cubic with a positive coefficient of ${x}^{3}$, we can deduce that the critical point corresponding to the smallest value of $x$ must be the maximum and that corresponding to the larger value must be the maximum. It is not possible to have any other possibility!

Nov 14, 2016

The max is at $\left(- 1 , 1\right)$
The min is at $\left(\frac{1}{3} , - \frac{5}{27}\right)$
The inflexion point is at $\left(- \frac{1}{3} , \frac{11}{27}\right)$

#### Explanation:

We have to calculate the first and second derivative.

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

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

$f ' \left(x\right) = 0$ when $x = \frac{1}{3}$ and $x = - 1$

So, we do a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$\frac{1}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\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}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{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}$$\uparrow$$\textcolor{w h i t e}{a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a}$$\uparrow$

So, we have a max at $x = - 1$ and a min at $x = \frac{1}{3}$

To determine the inflexion points, we calculate $f ' ' \left(x\right)$

$f ' ' \left(x\right) = 6 x + 2$

$f ' ' \left(x\right) = 0$ when $x = - \frac{1}{3}$

The inflexion point is at $x = - \frac{1}{3}$

Also, $f ' ' \left(- 1\right) = - 4 < 0$ which is a max

and $f ' ' \left(\frac{1}{3}\right) = 4 > 0$ which is a min

graph{x^3+x^2-x [-2.43, 2.436, -1.217, 1.215]}