`f(x)=3x^5-5x^3`
Find Increasing and decreasing intervals of a function

Solution:
Here, `f(x)=3x^5-5x^3`

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

`=d/(dx)(3x^5)-d/(dx)(5x^3)`

`=15x^4-15x^2`

---

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`

`=>15x^4-15x^2 = 0`

`=>15x^2(x^2-1) = 0`

`=>15x^2((x)^2-(1)^2) = 0`

`=>15x^2(x-1)(x+1) = 0`

`=>x^2 = 0" or "(x-1) = 0" or "(x+1) = 0`

`=>x^2 = 0" or "x = 1" or "x = -1`

`=>x = 0" or "x = 1" or "x = -1`

The solution is
`x = 0,x = 1,x = -1`

`:.` `x=-1`, `x=0` and `x=1`

---

Step-3: Use the critical points to determine intervals
There are total 3 critical points, So we have 4 intervals
`(-oo,-1),(-1,0),(0,1),(1,oo)`

---

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)``=15*(-2)^4-15*(-2)^2`

`=240-60`

`=180`` > 0`

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

2. For second interval `(-1,0)`, we choose `x=-0.5`

`f^'(-0.5)``=15*(-0.5)^4-15*(-0.5)^2`

`=0.9375-3.75`

`=-2.8125`` < 0`

`:.` Function is decreasing on `(-1,0)`

3. For third interval `(0,1)`, we choose `x=0.5`

`f^'(0.5)``=15*0.5^4-15*0.5^2`

`=0.9375-3.75`

`=-2.8125`` < 0`

`:.` Function is decreasing on `(0,1)`

4. For fourth interval `(1,oo)`, we choose `x=2`

`f^'(2)``=15*2^4-15*2^2`

`=240-60`

`=180`` > 0`

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

Interval	x-value	`f^'(x)`	increasing or decreasing
`(-oo,-1)`	`-2`	`f^'(-2)=180`` > 0`	f is increasing
`(-1,0)`	`-0.5`	`f^'(-0.5)=-2.8125`` < 0`	f is decreasing
`(0,1)`	`0.5`	`f^'(0.5)=-2.8125`` < 0`	f is decreasing
`(1,oo)`	`2`	`f^'(2)=180`` > 0`	f is increasing

So, function f(x) is increasing on `(-oo,-1),(1,oo)` and decreasing on `(-1,0),(0,1)`

---

---