find the intervals where the function is increasing and the intervals where it is decreasing

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

ok i found its derivative
which is (f1(x)=(x-1)x

my options are increasing a) (0,infinity) b) (-1,0) c(-infinity,0)U(1,infinity)

decreasing a) (0,1) b) (-infinity,0)U(1,infinity)

i cant seem to find the correct answers
because it is increasing from -infinitive to -1. then increasing from -1,0 decreasing from 0,1 and increasing again from (1, infinity)

what im i doing wrong?

You have the correct calculations, but did not organize in a way that helps you get the right answer.

"because it is A(increasing from -infinitive to -1). then B(increasing from -1,0), C(decreasing from 0,1)) and D(increasing again from (1, infinity))"

Statements A, B C and D are all correct.
Since A and B are both increasing, so can you not combine A and B?
Therefore you have two intervals that increase, and one single interval the decreases.

Post again if you still don't find the correct answer.

To determine the intervals where the function is increasing and decreasing, we need to analyze the sign of the derivative.

The derivative of the function f(x) = (x^3/3) - (x^2/2) is f'(x) = (x-1)x.

To find the intervals where the function is increasing, we need to identify where the derivative is positive, which means f'(x) > 0.

Similarly, to find the intervals where the function is decreasing, we need to identify where the derivative is negative, which means f'(x) < 0.

Now let's analyze the derivative:

f'(x) = (x-1)x

We can determine its sign by observing the values of x.

1. Set f'(x) = 0 to find the critical points:

(x-1)x = 0

The derivative is equal to zero when x = 0 and x = 1.

2. Test the intervals between the critical points and beyond:

a) For x < 0: Select a value, e.g., -2, plug it into the derivative.
f'(-2) = (-2-1)(-2) = (-3)(-2) = 6
Since f'(-2) > 0, the derivative is positive. Thus, the function is increasing on the interval (-∞, -1).

b) For 0 < x < 1: Select a value, e.g., 0.5, plug it into the derivative.
f'(0.5) = (0.5-1)(0.5) = (-0.5)(0.5) = -0.25
Since f'(0.5) < 0, the derivative is negative. Thus, the function is decreasing on the interval (0,1).

c) For x > 1: Select a value, e.g., 2, plug it into the derivative.
f'(2) = (2-1)(2) = (1)(2) = 2
Since f'(2) > 0, the derivative is positive. Thus, the function is increasing on the interval (1, ∞).

So, to summarize, the function is increasing on the intervals (-∞, -1) and (1, ∞), and it is decreasing on the interval (0, 1).