To determine the extreme value for a function, minimum or maximum, we have to differentiate the function,which, in this case, is:

f'(x)= (x^3-3x)'

f'(x)= 3x^2 - 3

Now, we'll calculate the root of the first derivative, this value being the value for the function f has an extreme value.

3x^2 - 3 = 0

We'll factorize:

3(x^2-1)=0

We'll solve the difference of squares:

3(x-1)(x+1)=0

We'll put each factor of the product as being zero.

x-1=0

x=1

x+1=0

x=-1

So, the extreme values of the function are:

f(1) = 1^3-3

**f(1) = -2**

f(-1) = -1+3

**f(-1) = 2**

To find the extreme values of f(x) =x^3-3x.

Solutions:

f(x) = x^3-3x = x(x^2-1) = x(x^2-1) = (x+1)(x)(x-1).

For x>1, f(x) positive is increasing and goes unbounded as x-->infinity.

For x=1, f(x) = 0.

For 0<x<1, f(x) is <0 or negative,

For x= 0, f(x) = 0.

For -1< x < 0, f(x) > 0 or positive.

For x =-1 , f(x) = 0 and

For X < -1, f(x) is negative and goes to minus infinity.

The local extreme values of the function is got by differentiating f(x) with respect to x setting the f'(x) = 0 and solving for x=c. Again if f"(c) < 0, then f(c) is mximum for x= c. If f"(c) > 0, then f'(c) is the minimum for x = c.

f(x) = x^3-3x. Differentiating both sides, setting f'(x) = 0,

3x^2-3 = 0. Or

x^2-1 = 0. Or

x^2 = 1. Or x = 1 Or x = -1.

f''(x) = (3x^2-3)' = 6x.

f''(1) =6*1 is positive. So for x =1. f(1) = 1^3-3*1 = -3 is a local minimum.

f"(-1) = 6(-1) = -1 . So f(-1) = 3(-1)^2-3 = 0 is the local maximum.