# How do you find the local maximum and minimum values of  f(x)=x^3 + 6x^2 + 12x -1 using both the First and Second Derivative Tests?

May 3, 2017

There are no minimum or maximum, only a point of inflection at $\left(- 2 , - 9\right)$

#### Explanation:

Let's calculate the first and second derivatives

$f \left(x\right) = {x}^{3} + 6 {x}^{2} + 12 x - 1$

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

$= 3 \left({x}^{2} + 4 x + 4\right)$

$= 3 \left(x + 2\right) \left(x + 2\right)$

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

and

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

The critical point is when $f ' \left(x\right) = 0$

That is,

$x + 2 = 0$

$x = - 2$

Let's build a 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}$$- 2$$\textcolor{w h i t e}{a a 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 a}$$+$$\textcolor{w h i t e}{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}$↗$\textcolor{w h i t e}{a a a a a a}$↗

Let's look at $f ' ' \left(x\right)$

$f ' ' \left(x\right) = 0$ when $x + 2 = 0$

That is when $x = - 2$

There is a point of inflection at $\left(- 2 , - 9\right)$

Let's do a 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 a}$$- 2$$\textcolor{w h i t e}{a a 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 a}$$-$$\textcolor{w h i t e}{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}$$\cap$$\textcolor{w h i t e}{a a a a a a}$$\cup$ graph{x^3+6x^2+12x-1 [-38.27, 26.68, -22.07, 10.37]}