Identify the open intervals on which the function f(x)=3x-x^3 is increasing or decreasing?

The sign of the 1st derivative tells us about the monotony
of a function.

If the 1st derivative is strictly positive,
then the function is strictly increasing and if the 1st derivative is strictly negative,
then the function is strictly decreasing.

We'll
differentiate the function with respect to x:

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

We'll detemrine the roots of
f'(x):

3-3x^2 = 0

1 - x^2
=0

The difference of two squares will return the
product:

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

We'll
cancel each factor:

1-x=0
=>x=1

1+x=0 =>
x=-1

The derivative is positive over (-1,1) and it is
negative over the intervals (-`oo` ,-1) and (1,`oo`
)

Therefore, the function is decreasing over
the intervals (-`oo` ,-1) and (1,`oo` ) and it is increasing over the interval
(-1,1).