Decreasing Intervals - CLEP Calculus

The function is decreasing where . To determine where this is happening, differentiate the function and find where . This will split the function into intervals where it is either increasing or decreasing.

To determine where this equals zero, factor:

this has solutions for .

Test a point in each region to determine if it is increasing or decreasing within these bounds:

positive/increasing

negative/decreasing

negative/decreasing

positive/increasing

The function is increasing where . To determine the regions where this is true, first take the derivative of :

.

To figure out where this is less than zero, factor and set it equal to zero. This will split the function into intervals where we can test points.

This has solutions at .

Test a point in each region to see if the function is increasing or decreasing:

positive/increasing

negative/decreasing

positive/increasing

To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, take the derivative:

Set equal to 0 and solve:

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of -6, 0, and 2.

Since the only value that is negative is when x=0, the interval is only decreasing on the interval that includes 0. Therefore, our answer is:

To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, take the derivative:

Set equal to 0 and solve:

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of 0, 2, and 10.

Since the only value that is negative is when x=0, the interval is only decreasing on the interval that includes 2. Therefore, our answer is:

To find the an increasing or decreasing interval, we need to find out if the first derivative is positive or negative on the given interval. So, find by decreasing each exponent by one and multiplying by the original number.

Next, we can find and and see if they are positive or negative.

Both are negative, so the slope of the line tangent to is negative, so is decreasing.

Recall that a function is increasing at a point if its first derivative is positive, and a function is decreasing if its first derivative is negative at that point. Therfore, we should start by finding f'(x). However, I will start by combining like terms and putting f(x) in standard form:

Next, plug in each of our endpoints to see what the sign of f'(x) is.

So f'(x) is positive on the given interval, so we know that f(x) is increasing on the given interval.

The first step is to find the first derivative.

We can factor a out to get

.

Now we need to solve for when to get the critical points. Notice how factoring the 2 made the expression a little easier to simplify.

The final step is to try points in all the regions to see which range gives a negative value for .

If we plugin in a number from the first range into , i.e , we get a positive number.

From the second range, , we get a negative number.

From the third range, , we get a positive number.

So the second range gives us values where the function is decreasing because is negative during that range, so is the answer.

The function is decreasing when the first derivative is negative. We first find when the derivative is zero. To find the derivative, we apply the quotient rule,

.

Therefore the derivative is zero at . To find when it is negative plug in test points on each of the three intervals created by these zeros.

For instance,

.

Hence the function is decreasing on

.

A function is strictly decreasing on an inverval if, for any in the interval, (i.e the slope is always less than zero)

Interval E is the only interval on which the function shows this property.

is decreasing on intervals where .

First, differentiate .

Then, find the values for x for which the derivative is negative by solving

.

Next, test the intervals.

Test them by substituting values for x:

Substitute -2,

.

The function is increasing on this interval since the derivative is positive on this interval.

Substitute 0,

.

The function is deecreasing on this interval since the derivative is negative on this interval.

Substitute 2,

.

The function is increasing on this interval since the derivative is positive on this interval.

Thus, is the only interval on which the function is decreasing.

To determine where the function is decreasing, differentiate it:

What we are interested in are the points where . To determine these points, factor the equation:

this has solutions at

This splits the graph into 4 regions, and we can test points in each to determine if is greater than or less than 0. If it is less than zero, the function is decreasing.

negative/decreasing

positive/increasing

negative/decreasing

positive/increasing

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 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 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 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 .