# What are the critical points of f(x) = x^3 − 12x + 7?

Jan 7, 2016

The critical points are:

MIN: (2;-9)
MAX: (-2;23)

#### Explanation:

The critical points are the x values where:

$f ' \left(x\right) = 0$

We obtain:

$f ' \left(x\right) = 3 {x}^{3 - 1} - 12 {x}^{1 - 1} + 0 = 3 {x}^{2} - 12 = 3 \cdot \left({x}^{2} - 4\right) =$
$= 3 \cdot \left(x + 2\right) \left(x - 2\right)$

Then:

$f ' \left(x\right) = 0 \iff 3 \cdot \left(x + 2\right) \left(x - 2\right) = 0$

a multipication is zero if the factors are zero:

$x + 2 = 0$

${x}_{1} = - 2$

$x - 2 = 0$

${x}_{2} = + 2$

Now, to find if they are min or max you have to evaluate $f ' \left(x\right) > 0$

$f ' \left(x\right) > 0$ for x in ]-oo;-2[ uu ]2;+oo[

then

$f \left(x\right)$ is growing for x in ]-oo;-2[ uu ]2;+oo[

$f ' \left(x\right) < 0$ for x in ]-2;2[

then

$f \left(x\right)$ is decreasing for x in ]-2;2[

then
for $x = - 2$ we have a MAX
for $x = 2$ we have a MIN

$f \left(- 2\right) = - 8 + 24 + 7 = 23$

$f \left(2\right) = 8 - 24 + 7 = - 9$