# If #f(x) = 2x^3 + 3x^2 - 180x#, how do I find the intervals on which f is increasing and decreasing?

Hello,

- You have to calculate the derivative :

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To find the intervals on which ( f(x) ) is increasing or decreasing:

- Calculate the derivative of ( f(x) ), denoted as ( f'(x) ).
- Set ( f'(x) ) equal to zero and solve for ( x ). These are the critical points.
- Test the intervals between the critical points by evaluating ( f'(x) ) at a point within each interval.
- If ( f'(x) > 0 ) in an interval, then ( f(x) ) is increasing on that interval. If ( f'(x) < 0 ), then ( f(x) ) is decreasing on that interval.

So, for ( f(x) = 2x^3 + 3x^2 - 180x ):

- ( f'(x) = 6x^2 + 6x - 180 ).
- Set ( f'(x) = 0 ): ( 6x^2 + 6x - 180 = 0 ). Solve for ( x ) to find the critical points.
- After finding the critical points, evaluate ( f'(x) ) at a point within each interval.
- Determine the intervals where ( f'(x) > 0 ) (increasing) and ( f'(x) < 0 ) (decreasing).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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