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

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A function f(x) is increasing when f'(x) is positive and it is decreasing when f'(x) is negative.

f(x) = 3x - x^3

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

First determine where f'(x) = 0

3 - 3x^2 = 0

=> 1 - x^2 = 0

=> x^2 = 1

=> x = -1 and x = 1

We have the intervals (-inf., -1), (-1, 1) and (1, inf.)

In (-inf., -1), f'(x) is seen to be less than 0. Hence the unction is decreasing here. Similarly, the function is decreasing in (1, inf.). The function is increasing is (-1, 1).

**The function is increasing in the interval (-1, 1) and decreasing elsewhere.**

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).**