To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative. 

The first derivative of the function is

and we used the following rule:

Now, we find the point(s) at which the first derivative equals zero - the critical value(s):

Now, we make our intervals on which we see whether the first derivative is positive or negative:

On the first interval, the first derivative is negative, while on the second interval it is positive. Thus, the first interval is the one where the function is decreasing. 

To find when a function is decreasing, we must first find where the critical points of the function are. Since we are given the derivative , we start by first setting the derivative equal to  and solving for .

This is our critical point. To evaluate where the function is decreasing, we must check the sign of the derivative on both sides of the critical point. Because , we can check the left side of the critical point by plugging  into .

Because , the function is decreasing on the left side of the critical point, on the interval .

Now we must check the right side of the critical point. Because , we can check the right side of the critical point by plugging  into .

Because , the function is increasing on the right side of the critical point, meaning the only interval on which the function is decreasing is .

To determine the intervals on which the function is decreasing, we must find the intervals on which the function's first derivative is negative. To do this, we must find the first derivative and the critical value(s) at which the first derivative is equal to zero:

The derivative was found using the following rule:

Now, setting the first derivative equal to zero, we get

So, now we can make our intervals to be analyzed (is the first derivative positive or negative on the interval?), in which c is the upper and lower bound:

Note that at c the first derivative is neither positive nor negative.

On the first interval, the first derivative is always negative, so the function is always decreasing on this interval. On the second interval, the first derivative is always positive, therefore the function is increasing on this interval. 

We are concerned with the interval where the function is decreasing, so  is our answer.

To determine where the function is decreasing, we must find the intervals on which the first derivative of the function is negative. 

First, we must find the first derivative:

The first derivative was found using the following rule:

Now, we set the first derivative equal to zero and solve for the critical values - values at which the first derivative is equal to zero:

Now, using the critical values, we create the intervals in which we see whether the first derivative is positive or negative:

On the first interval, the derivative is positive, on the second it is negative, and on the third interval it is positive (we simply plug in any point in the interval into the first derivative function and check the sign). 

So, the function is decreasing on the second interval .

To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

First, we find the first derivative:

It was found using the following rule:

Now, we must find the critical values - values at which the first derivative is equal to zero:

Now, using the critical values as upper and lower bounds, we create the intervals on which we determine the sign of the first derivative:

On the first interval, the first derivative is always positive, and on the second interval, the first derivative is always positive. (Simply plug in any point in the interval into the first derivative function and check the sign.) Thus, the function is never decreasing.

To find the intervals where the function decreases, apply the first derivative test. Find the derivative, set equal to , and solve to find local extrema.

So  or .

Next, test points in each of the intervals delineated by the potential local extrema. For example:

, so the function increases to the left of .

, so the function decreases on the interval

, so the function increases to the right of .

To find decreasing intervals, you must find when the first derivative is less than 0. Differentiate using the power rule:

Thus,

Since 2 is never negative, our first derivative is never negative. Therefore, our function is never decreasing.

To find the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

So, the first derivative of the function is equal to

and was found using the following rule:

Next, we must find the critical values, at which the first derivative is equal to zero:

Now, we make the intervals, using c as our upper and lower bound:

To determine whether the first derivative is positive or negative each interval, simply plug in any number on the interval into the first derivative function. On the first interval, the first derivative is positive, while on the second interval, the first derivative is negative. Our answer is therefore .

To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative is negative.

The first derivative of the function is 

and was found using the following rule:

Now, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we can make our intervals:

Note that at the bounds of the intervals the first derivative is neither positive nor negative.

To determine the sign of the first derivative, simply plug in any point on the interval into the first derivative function: on the first interval, the first derivative is positive, on the second interval it is negative, and on the third interval it is positive. Thus, our answer is .

We need to set the derivative equal to zero.  Then, we can simply input values to the left and right of the roots to see their sign.  If the output is negative, the function is increasing on that range.  

The derivative is negative only in the region between these roots.  Therefore, the initial function is decreasing only in between these two values.