`f(x)=x^3+3x^2-9x+7`
Find Increasing and decreasing intervals of a function

Solution:
Here, `f(x)=x^3+3x^2-9x+7`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(x^3+3x^2-9x+7)`

`=d/(dx)(x^3)+d/(dx)(3x^2)-d/(dx)(9x)+d/(dx)(7)`

`=3x^2+6x-9+0`

`=3x^2+6x-9`

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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+6x-9 = 0`

`=>3(x^2+2x-3) = 0`

`=>3(x^2-x+3x-3) = 0`

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

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

`=>(x-1) = 0" or "(x+3) = 0`

`=>x = 1" or "x = -3`

The solution is
`x = 1,x = -3`

`:.` `x=-3` 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,-3),(-3,1),(1,oo)`

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Step-4: Determine if the function is increasing or decreasing in each interval
1. For first interval `(-oo,-3)`, we choose `x=-4`

`f^'(-4)``=3*(-4)^2+6*(-4)-9`

`=48-24-9`

`=15`` > 0`

`:.` Function is increasing on `(-oo,-3)`

2. For second interval `(-3,1)`, we choose `x=0`

`f^'(0)``=3*0^2+6*0-9`

`=0+0-9`

`=-9`` < 0`

`:.` Function is decreasing on `(-3,1)`

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

`f^'(2)``=3*2^2+6*2-9`

`=12+12-9`

`=15`` > 0`

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

Interval	x-value	`f^'(x)`	increasing or decreasing
`(-oo,-3)`	`-4`	`f^'(-4)=15`` > 0`	f is increasing
`(-3,1)`	`0`	`f^'(0)=-9`` < 0`	f is decreasing
`(1,oo)`	`2`	`f^'(2)=15`` > 0`	f is increasing

So, function f(x) is increasing on `(-oo,-3),(1,oo)` and decreasing on `(-3,1)`

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