# What are the local extrema, if any, of f (x) = x^3-12x+2 ?

Nov 9, 2015

The function has 2 extrema:

${f}_{\max} \left(- 2\right) = 18$ and ${f}_{\min} \left(2\right) = - 14$

#### Explanation:

We have a function: $f \left(x\right) = {x}^{3} - 12 x + 2$

To find extrema we calculate derivative

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

The first condition to find extreme points is that such points exist only where $f ' \left(x\right) = 0$

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

3(x^2-4)=0)

$3 \left(x - 2\right) \left(x + 2\right) = 0$

$x = 2 \vee x = - 2$

Now we have to check if the derivative changes sign at the calcolated points:

graph{x^2-4 [-10, 10, -4.96, 13.06]}

From the graph we can see that $f \left(x\right)$ has maximum for $x = - 2$ and minimum for $x = 2$.

Final step is to calculate the values $f \left(- 2\right)$ and $f \left(2\right)$