# Increasing and decreasing functions: find the intervals where f(x)is increasing or decreasing. f(x)=x-x^2. I believe the answer is increasing (1/2,inf)so far I have x-x^2 1-2x=0 -2x=-1 x=1/2 and I...

Increasing and decreasing functions: find the intervals where f(x)is increasing or decreasing. f(x)=x-x^2. I believe the answer is increasing (1/2,inf)

so far I have x-x^2

1-2x=0

-2x=-1

x=1/2

and I figured that meant the function was increasing at (1/2,infinity)

I'm not sure I'm understanding this right, so I want to know if what I've done is right, or how it's supposed to be done.

thanks.

### 2 Answers | Add Yours

`f(x) = x-x^2`

For any function maximum or minimum points can be found by the first derivative.

`(df(x))/dx = 1-2x`

At maximum and minimum points `(df(x))/dx = 0 `

`(df(x))/dx = 0`

`1-2x = 0`

` x = 1/2`

So at x = 1/2 there is a point of a graph which is a maximum of minimum.

If `(d^2f(x))/(dx^2) < 0` then the point is a maximum point and If `(d^2f(x))/(dx^2) > 0` then the point is a minimum point.

`(d^2f(x))/(dx^2) = -2 <0`

Therefore at x=1/2 we have a maximum point. So before x = 1/2 the function is increasing and after x = 1/2 the function is decreasing.

f(x) is increasing when `x in (-oo,1/2)`

f(x) is decreasing when `x in (1/2,-oo)`

For further clarification you can see the graph of `f(x) = x-x^2` below.

**Sources:**

Thank you, that was one of the answers that I had gotten, It's good to see that I am doing something rightish =D