`f(x)=x^3-3x+2`
Find Increasing and decreasing intervals of a function

Solution:
Here, `f(x)=x^3-3x+2`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(x^3-3x+2)`

`=d/(dx)(x^3)-d/(dx)(3x)+d/(dx)(2)`

`=3x^2-3+0`

`=3x^2-3`

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Step-2: Find the critical points of the derivative function
To find critical points, set `f^'(x)=0` and then solve for x

`f^'(x)=0`

`=>3x^2-3 = 0`

`=>3x^2 = 3`

`=>x^2 = 3/3`

`=>x^2 = 1`

`=>x = +- 1`

The solution is
`x = 1,x = -1`

`:.` `x=-1` and `x=1`

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Step-3: Use the critical points to determine intervals
There are total 2 critical points, So we have 3 intervals
`(-oo,-1),(-1,1),(1,oo)`

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Step-4: Determine if the function is increasing or decreasing in each interval
1. For first interval `(-oo,-1)`, we choose `x=-2`

`f^'(-2)``=3*(-2)^2-3`

`=12-3`

`=9`` > 0`

`:.` Function is increasing on `(-oo,-1)`

2. For second interval `(-1,1)`, we choose `x=0`

`f^'(0)``=3*0^2-3`

`=0-3`

`=-3`` < 0`

`:.` Function is decreasing on `(-1,1)`

3. For third interval `(1,oo)`, we choose `x=2`

`f^'(2)``=3*2^2-3`

`=12-3`

`=9`` > 0`

`:.` Function is increasing on `(1,oo)`

Interval	x-value	`f^'(x)`	increasing or decreasing
`(-oo,-1)`	`-2`	`f^'(-2)=9`` > 0`	f is increasing
`(-1,1)`	`0`	`f^'(0)=-3`` < 0`	f is decreasing
`(1,oo)`	`2`	`f^'(2)=9`` > 0`	f is increasing

So, function f(x) is increasing on `(-oo,-1),(1,oo)` and decreasing on `(-1,1)`

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