In this section, we use the derivative to determine intervals on which a given function is increasing or decreasing. We will also determine the local extremes of the function.
1 Increasing and Decreasing Functions
Similarly, is called decreasing on an interval if given any two numbers, and in such that , we have .
The derivative is used to determine the intervals where a function is either increasing or decreasing. The following theorem is a direct consequence of the cornerstone, Mean Value Theorem (section 3.3).
If on , then is increasing on and,
if on , then is decreasing on .
- Proof
- (Increasing case): Let and be in the interval with . Since is
differentiable on the closed interval , it is also continuous there. Hence, we can
apply the MVT on to on and conclude that there is a number between and
such that Since on the interval (by hypothesis), we know that . We also know
that since . Therefore, we must have which is the same as . We can now see
that if on and if and are any two values in with then we have . This means
that is increasing on .
The decreasing case is proved similarly.
We now use this theorem to determine intervals on which a given function is either increasing or decreasing.
The graph of a function of this form is a straight line with slope, . If , then is increasing on the interval and if , then it is decreasing on .
The derivative of this polynomial is . Since for all values of ,
on the interval . Thus is an increasing function on .
The derivative of this polynomial is . If , then which means that is decreasing on the interval . Similarly, if , then and so is increasing on the interval .
Is increasing or decreasing on the interval ?
Is increasing or decreasing on the interval ?
The domain of is and since for all . By the increasing/decreasing theorem, is decreasing on its domain, .
Is increasing or decreasing on the interval ?
Is increasing or decreasing on the interval ?
The domain of is and By the increasing/decreasing theorem, is increasing on its domain, .
Is increasing or decreasing on the interval ?
Is increasing or decreasing on the interval ?
The function
- (a)
- is even, i.e.,
- (b)
- is continuous on its domain, ,
- (c)
- has an infinite discontinuity at .
- (d)
- is differentiable (for ) and
To see the last property, let and use the chain rule: This derivative can now be used to determine the intervals of increase and decrease. Observe that By the increasing/decreasing theorem, we can conclude that :
- (a)
- is increasing on the interval , and
- (b)
- is decreasing on the interval .
This can be seen in the graph below.
Is increasing or decreasing on the interval ?
According to the theorem, we must determine where is positive and where is negative. To do this, it is often easiest to first determine where or is undefined. In this example, which exists for all . We solve the equation which yields and hence, or . Note that these are the critical numbers of . These two -values break the real number line into three open intervals: and . On each of these intervals, will be either strictly positive or strictly negative. To determine which, we will use test points. For the interval, we will use as the test point (any number in the interval is acceptable). Then we consider the derivative at this point: . This means that for every value in the interval and by the theorem, is increasing on this interval.
Next, for the interval , we will use as the test point, and we have . This means that for every value in the interval and by the theorem, is decreasing on this interval.
Finally, for the interval , we will use as the test point and we have . This means that for every value in the interval and by the theorem, is increasing on this interval.
Thus is increasing on the intervals and , and it is decreasing on the interval .
According to the theorem, we must determine where is positive and where is negative. To do this, it is often easiest to first determine where or is undefined. By the product rule, which exists for all . There is one solution to the equation, and that is . Note that for any value of .
This critical number breaks the real number line into two open intervals: and . On each of these intervals, will be either strictly positive or strictly negative. To determine which, we will use test points. For the interval, we will use as the test point (any number in the interval is acceptable). Then we consider the derivative at this point: . This means that for every value in the interval and by the theorem, is decreasing on this interval.
For the interval , we will use as the test point and we have . This means that for every value in the interval and by the theorem, is increasing on this interval.
Thus is decreasing on the interval and it is increasing on the interval .
The work that was done in the previous example can actually give us slightly more information about . We can determine the local extremes of .
Critical numbers can help us find the location and the nature of local extremes and the next theorem tells us how.
In a prior example, we found that the critical numbers for are and . We also noted that changed sign from positive to negative at and from negative to positive at . Hence, by the First Derivative Test, has a local maximum at and a local minimum at .
If there are none, type “none”.
has a local maximum at
has a local minimum at
If there are none, type “none”.
has a local maximum at
has a local minimum at
If there are none, type “none”.
has a local maximum at
has a local minimum at
If there are none, type “none”.
If there is more than one local extreme, list them in ascending order.
has a local maximum at
has a local minimum at
and at
If there are none, type “none”.
has a local maximum at
has a local minimum at
We begin by finding the critical numbers of . By the product and chain rules, The derivative exists for all . Setting the derivative equal to zero gives The first equation has no solutions, since raised to any power is strictly positive and the second equation has one solution, . This one critical number breaks the real number line into two intervals, and . By the Intermediate Value Theorem, the derivative will not change sign on these intervals. We choose a test point from each interval to plug into to determine its sign on the corresponding interval. From , we will choose the and from we will choose . Thus, and so is negative on the interval . Similarly, and so is positive on the interval . This information is summarized in the sign chart below.
Applying the increasing/decreasing theorem, we can conclude that is decreasing on the interval and increasing on the interval .
As for the local extremes, at the critical number , the derivative changes sign from negative to positive, so by the First Derivative Test, has a local minimum at . There are no other critical numbers, so there are no other local extremes for this function.
If there are none, type “none”.
has a local maximum at
has a local minimum at
If there are none, type “none”.
has a local maximum at
has a local minimum at
If there are none, type “none”.
has a local maximum at
has a local minimum at
First, recall that the Increasing/Decreasing Theorem states that is increasing on intervals where and is decreasing on intervals where . Such intervals can be determined from the graph of by noting when it is above or below the -axis. This graph is above the -axis on the interval and below the -axis on the intervals and . When the graph of is above the -axis, then and hence, is increasing. Likewise, when the graph of is below the -axis, then and is decreasing. We can now interpret the graph of to state that is increasing on the interval and is decreasing on the intervals and .
is increasing on the interval(s):
The local extremes of are: