Question

I don't know how to solve it.

Answer 1

Step-by-step explanation:

Take the first derivative

$\frac{d}{dx} ( {x}^{3} - 3x)$

$3 {x}^{2} - 3$

Set the derivative equal to 0.

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

$3 {x}^{2} = 3$

${x}^{2} = 1$

$x = 1$

or

$x = - 1$

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

$( - 1) {}^{3} - 3( - 1) = 2$

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

${1}^{3} - 3(1)$

$1 - 3 = - 2$

So the local min occurs at -2, at x=1