Finding intervals where function is increasing or decreasing

[919_is]

Find interval(s) where function is Decreasing:

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

f'(x) = 2 - 2x^-1/3

Set f'(x) equal to zero:

(2x^-1/3)/2 = 2/2

1 = 1x^1/3

x = 1


Therefore, only one critical number and partition number: x = 1

However, when I construct my line chart for f'(x), the function increases at (-infinity, 0), decreases at (0,1), and increases again at (1, infinity). My question is, where did the zero (0) become a partition number in my line chart when I only get one critical value? Where did zero come from?

Please advise, and thank you in advance.

 

[biffboy]

To find interval where decreasing just solve f '(x) < 0

 

[HallsofIvy]

Find interval(s) where function is Decreasing:

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

f'(x) = 2 - 2x^-1/3

Set f'(x) equal to zero:

(2x^-1/3)/2 = 2/2

1 = 1x^1/3

x = 1


Therefore, only one critical number and partition number: x = 1

However, when I construct my line chart for f'(x), the function increases at (-infinity, 0), decreases at (0,1), and increases again at (1, infinity). My question is, where did the zero (0) become a partition number in my line chart when I only get one critical value? Where did zero come from?

Please advise, and thank you in advance.

If you found only one critical point, perhaps you are not clear on what a "critical point" is.

A critical point is where the derivative is 0 OR where the derivative does not exist. The derivative is 0 at x= 1 and does not exist (because of that negative power on x) at x= 0.