`f(x)=0.2x^5+1.25x^4+2x^3+2`
Find Increasing and decreasing intervals of a function

Solution:
Here, `f(x)=0.2x^5+1.25x^4+2x^3+2`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(0.2x^5+1.25x^4+2x^3+2)`

`=d/(dx)(0.2x^5)+d/(dx)(1.25x^4)+d/(dx)(2x^3)+d/(dx)(2)`

`=x^4+5x^3+6x^2+0`

`=x^4+5x^3+6x^2`

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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`

`=>x^4+5x^3+6x^2 = 0`

`=>x^2(x^2+5x+6) = 0`

`=>x^2(x^2+2x+3x+6) = 0`

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

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

`=>x^2 = 0" or "(x+2) = 0" or "(x+3) = 0`

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

`=>x = 0" or "x = -2" or "x = -3`

The solution is
`x = 0,x = -2,x = -3`

`:.` `x=-3`, `x=-2` and `x=0`

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Step-3: Use the critical points to determine intervals
There are total 3 critical points, So we have 4 intervals
`(-oo,-3),(-3,-2),(-2,0),(0,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)``=(-4)^4+5*(-4)^3+6*(-4)^2`

`=256-320+96`

`=32`` > 0`

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

2. For second interval `(-3,-2)`, we choose `x=-2.5`

`f^'(-2.5)``=(-2.5)^4+5*(-2.5)^3+6*(-2.5)^2`

`=39.0625-78.125+37.5`

`=-1.5625`` < 0`

`:.` Function is decreasing on `(-3,-2)`

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

`f^'(-1)``=(-1)^4+5*(-1)^3+6*(-1)^2`

`=1-5+6`

`=2`` > 0`

`:.` Function is increasing on `(-2,0)`

4. For fourth interval `(0,oo)`, we choose `x=1`

`f^'(1)``=1^4+5*1^3+6*1^2`

`=1+5+6`

`=12`` > 0`

`:.` Function is increasing on `(0,oo)`

Interval	x-value	`f^'(x)`	increasing or decreasing
`(-oo,-3)`	`-4`	`f^'(-4)=32`` > 0`	f is increasing
`(-3,-2)`	`-2.5`	`f^'(-2.5)=-1.5625`` < 0`	f is decreasing
`(-2,0)`	`-1`	`f^'(-1)=2`` > 0`	f is increasing
`(0,oo)`	`1`	`f^'(1)=12`` > 0`	f is increasing

So, function f(x) is increasing on `(-oo,-3),(-2,0),(0,oo)` and decreasing on `(-3,-2)`