WEBVTT
00:00:01.460 --> 00:00:09.990
In this video, weβre going to learn how to find the intervals over which a function is increasing, decreasing, or constant.
00:00:11.030 --> 00:00:20.570
We say that a function is increasing when the value of the function π of π₯ increases as the value of π₯ increases.
00:00:21.060 --> 00:00:24.040
This will result in a graph that slopes upwards.
00:00:24.400 --> 00:00:31.330
And so the slope of the graph of a function over an interval during which it is increasing must be positive.
00:00:31.730 --> 00:00:40.330
We can, conversely, say that a function will be decreasing if the value of π of π₯ decreases as the value of π₯ increases.
00:00:41.000 --> 00:00:47.660
It then follows that if a function is decreasing over that interval, the slope of its graph will be negative.
00:00:48.190 --> 00:00:55.520
For a function to be strictly increasing or strictly decreasing, there can be no flat bits on the graph of that function at all.
00:00:55.890 --> 00:01:03.870
If we have a flat piece of graph, in other words, a horizontal line, we say that the function is constant over this interval.
00:01:04.700 --> 00:01:10.610
Now, of course, we might not necessarily be given the graph of the function, so we can generalize these ideas.
00:01:10.960 --> 00:01:18.560
A function is increasing if when π₯ two is greater than π₯ one, π of π₯ two is greater than or equal to π of π₯ one.
00:01:18.970 --> 00:01:24.440
Then itβs strictly increasing if π of π₯ two is just greater than π of π₯ one.
00:01:25.150 --> 00:01:32.890
If when π₯ two is greater than π₯ one, π of π₯ one is equal to π of π₯ two, the function is constant over that interval.
00:01:33.510 --> 00:01:39.390
In a similar way, we form definitions for functions that are decreasing and strictly decreasing.
00:01:40.680 --> 00:01:48.140
Now, throughout this video, weβre also going to use interval notation to describe the intervals of increase and decrease.
00:01:48.240 --> 00:01:49.620
So letβs recall these.
00:01:50.480 --> 00:01:52.280
π
is the set of real numbers.
00:01:52.600 --> 00:01:57.560
These are the numbers that we use most often, and they include rational numbers and irrational numbers.
00:01:57.940 --> 00:02:01.870
But they donβt include imaginary numbers or positive or negative β.
00:02:03.180 --> 00:02:09.880
Then square brackets or parentheses describe a set of values when we do want to include the end values.
00:02:11.630 --> 00:02:17.330
And then we use the round brackets or parentheses when we donβt want to include the end values on our interval.
00:02:18.000 --> 00:02:28.520
Weβre now going to consider a number of examples of using graphs to establish intervals of increase or decrease and also how weβre going to find these by using the equations.
00:02:30.330 --> 00:02:32.550
The graph of a function is given below.
00:02:33.210 --> 00:02:36.590
Which of the following statements about the function is true?
00:02:37.200 --> 00:02:40.560
Is it (A) the function is decreasing on the set of real numbers?
00:02:41.150 --> 00:02:44.550
Is it (B) the function is constant on the set of real numbers?
00:02:45.410 --> 00:02:52.310
(C) The function is increasing on the left-open right-closed interval from negative β to zero.
00:02:52.780 --> 00:02:56.330
Is it (D) the function is increasing on the set of real numbers?
00:02:56.700 --> 00:03:02.720
Or (E) the function is constant on the left-open right-closed interval from negative β to zero.
00:03:03.440 --> 00:03:11.330
Letβs begin by recalling what the words decreasing, increasing, and constant tell us about the graph of a function.
00:03:11.780 --> 00:03:20.540
If a function π of π₯ is decreasing over some interval, then the value of π of π₯ decreases as the value of π₯ increases.
00:03:20.800 --> 00:03:25.410
In terms of the graph, we can say that the graph will slope downwards over that interval.
00:03:25.800 --> 00:03:30.000
The opposite is true if a function is increasing over some interval.
00:03:30.370 --> 00:03:34.840
As the value of π₯ increases, the value of the function also increases.
00:03:35.240 --> 00:03:37.540
And then this looks like the graph sloping upwards.
00:03:38.090 --> 00:03:44.910
Then if a function is constant, as the value of π₯ increases, the value of the function remains the same.
00:03:45.170 --> 00:03:48.320
And in terms of the graph, this looks like a horizontal line.
00:03:48.700 --> 00:03:54.700
And if we compare our graph to these three terms and these criteria, we see we have a horizontal line.
00:03:54.980 --> 00:03:57.190
So our function must be constant.
00:03:57.470 --> 00:04:03.330
So if we compare these to our options (A) through (E), we see weβre looking at (B) and (E).
00:04:03.810 --> 00:04:13.990
(B) says the function is constant on the set of real numbers, whereas (E) says the function is constant on the left-open right-closed interval from negative β to zero.
00:04:14.310 --> 00:04:16.170
So which of these are we going to choose?
00:04:16.510 --> 00:04:25.120
If we think about this notation, this is telling us that the function is constant for all values less than and including zero.
00:04:25.530 --> 00:04:34.860
And in fact, this is a subset of the set of real numbers which extends from negative β to positive β but doesnβt include those endpoints.
00:04:35.090 --> 00:04:40.660
If we look at the horizontal line representing our function, we see it has arrows at both ends.
00:04:41.010 --> 00:04:47.540
And so our line itself must also extend up to positive β and down to negative β.
00:04:47.970 --> 00:04:53.890
And so we can actually say that the correct answer is (B); the function must be constant on the set of real numbers.
00:04:55.960 --> 00:05:06.520
In our next example, weβll see how to use interval notation to describe whether a function is increasing, decreasing, or constant over particular intervals.
00:05:07.830 --> 00:05:13.640
Which of the following statements correctly describe the monotony of the function represented in the figure below?
00:05:14.360 --> 00:05:25.200
Is it (A) the function is increasing on the open interval five to eight, constant on the open interval negative one to five, and decreasing on the open interval negative two to negative one?
00:05:25.540 --> 00:05:36.530
Is it (B) the function is increasing on the open interval negative two to negative one, constant on the open interval negative one to five, and decreasing on the open interval five to eight?
00:05:36.820 --> 00:05:44.120
Is it (C) the function is increasing on the open interval five to eight and decreasing on the open interval negative two to five?
00:05:44.500 --> 00:05:52.110
Or (D) the function is increasing on the open interval negative two to five and decreasing on the open interval five to eight.
00:05:52.700 --> 00:05:58.820
So by reading the question, weβve probably inferred what we mean by the monotony of a function.
00:05:59.260 --> 00:06:05.660
The monotony of a function simply tells us if the function is increasing or decreasing.
00:06:06.400 --> 00:06:13.140
And of course, we recall that if a function is increasing over some interval, it has a positive slope.
00:06:13.470 --> 00:06:16.120
If itβs decreasing, it has a negative slope.
00:06:16.420 --> 00:06:19.570
And if itβs constant, well, thatβs a horizontal line.
00:06:19.940 --> 00:06:22.530
So letβs look at the graph of our function.
00:06:22.820 --> 00:06:25.260
We see it has three main sections.
00:06:25.570 --> 00:06:29.510
The first section is between negative two and negative one.
00:06:30.000 --> 00:06:37.940
Then the next section is between negative one and five, whilst the third section is between five and eight.
00:06:38.150 --> 00:06:40.550
So letβs consider each section in turn.
00:06:41.000 --> 00:06:46.440
We can see that the slope of the first part of our function must be positive.
00:06:46.740 --> 00:06:48.490
Itβs sloping upwards.
00:06:48.610 --> 00:06:53.790
We then have a horizontal line between π₯ equals negative one and five.
00:06:54.030 --> 00:06:57.220
And the third part of our graph has a negative slope.
00:06:57.440 --> 00:06:58.820
Itβs sloping downwards.
00:06:59.260 --> 00:07:05.600
Our function is therefore increasing for sometime, itβs constant, and then finally itβs decreasing.
00:07:05.950 --> 00:07:09.680
We need to decide the intervals over which each of these occur.
00:07:09.970 --> 00:07:14.680
It has a positive slope between π₯ equals negative two and negative one.
00:07:15.020 --> 00:07:19.540
And so we define this using the open interval negative two to negative one.
00:07:20.070 --> 00:07:22.640
We are not going to use a closed interval.
00:07:22.810 --> 00:07:26.170
We donβt really know whatβs happening at the endpoints of this interval.
00:07:26.470 --> 00:07:32.420
For instance, when π₯ is equal to negative one, the graph of our function has this sort of sharp corner.
00:07:32.890 --> 00:07:38.660
And so weβre going to leave π₯ equals negative two and π₯ equals negative one out of our interval.
00:07:39.040 --> 00:07:43.800
In a similar way, the function is constant over the open interval negative one to five.
00:07:44.200 --> 00:07:48.260
And itβs decreasing over the open interval five to eight.
00:07:48.320 --> 00:07:53.790
Once again, we donβt know whatβs really happening at those endpoints, but we do have sharp corners.
00:07:53.790 --> 00:07:57.880
And so we canβt say whether itβs increasing, decreasing, or constant.
00:07:58.110 --> 00:08:11.280
And so the correct answer is (B): the function is increasing on the open interval negative two to negative one, constant on the open interval from negative one to five, and decreasing on the open interval five to eight.
00:08:13.500 --> 00:08:21.030
In our next example, weβre going to look at how to identify increasing and decreasing regions from a reciprocal graph.
00:08:23.060 --> 00:08:25.070
The graph of a function is given below.
00:08:25.660 --> 00:08:28.790
Which of the following statements about the function is true?
00:08:29.360 --> 00:08:38.060
Is it (A) the function is increasing on the open interval negative β to zero and increasing on the open interval zero to β?
00:08:38.550 --> 00:08:45.510
Is it (B) the function is decreasing on the open interval negative β to negative five and negative five to β?
00:08:45.860 --> 00:08:53.880
Is it (C) the function is increasing on the open interval negative β to negative five and the open interval negative five to β?
00:08:54.110 --> 00:09:03.030
Or (D) the function is decreasing on the open interval negative β to zero and decreasing on the open interval zero to β.
00:09:03.780 --> 00:09:06.970
Each of the statements is regarding the monotony of the graph.
00:09:07.250 --> 00:09:13.070
Itβs asking us whether the graph is increasing or decreasing over given intervals.
00:09:13.430 --> 00:09:22.300
And so we recall that we can say that a function is increasing if its value for π of π₯ increases as the value for π₯ increases.
00:09:22.540 --> 00:09:25.900
In terms of the graph, weβd be looking for a positive slope.
00:09:26.160 --> 00:09:31.450
Then if a function is decreasing, its graph will have negative slope over that interval.
00:09:31.720 --> 00:09:33.700
And so letβs have a look at our graph.
00:09:34.000 --> 00:09:36.820
It appears to be the graph of a reciprocal function.
00:09:37.190 --> 00:09:39.900
And the graph has two asymptotes.
00:09:40.220 --> 00:09:45.860
We see that the π¦-axis, which is the line π₯ equals zero, is a vertical asymptote.
00:09:46.300 --> 00:09:52.330
And then we have a horizontal asymptote given by the line π¦ equals negative five.
00:09:53.050 --> 00:09:58.740
Now what this means is that the graph of our function will approach these lines, but it will never quite meet them.
00:09:59.440 --> 00:10:06.760
And this, in turn, means that the graph of our function will never quite become a completely horizontal or completely vertical line.
00:10:07.510 --> 00:10:11.110
And so letβs see whatβs happening as our value of π₯ increases.
00:10:11.620 --> 00:10:18.580
As we move from negative β to zero, the function π of π₯ increases.
00:10:18.800 --> 00:10:26.550
Its slope is always positive, and each value of π of π₯ is greater than the previous value of π of π₯.
00:10:26.950 --> 00:10:33.190
Then when we move from π₯ equals zero to positive β, the same happens.
00:10:33.420 --> 00:10:40.530
And so this means that the graph is increasing from negative β to zero and from zero to β.
00:10:40.530 --> 00:10:41.990
But whatβs happening at zero?
00:10:42.340 --> 00:10:46.250
Well, we see that the function canβt take a value of π₯ equals zero.
00:10:46.600 --> 00:10:51.830
And so the graph of our function approaches the line π₯ equals zero but never quite reaches it.
00:10:52.150 --> 00:11:07.190
We then use these round brackets or parentheses to show that the graph is increasing between π₯ equals negative β and zero and between π₯ equals zero and β but that we donβt want to include the end values in these statements.
00:11:07.360 --> 00:11:12.950
Notice that we donβt include negative β and β because we canβt really define that number.
00:11:13.300 --> 00:11:23.580
And so the correct answer must be (A), the function is increasing on the open interval negative β to zero and increasing on the open interval zero to β.
00:11:25.510 --> 00:11:31.620
Weβre now going to consider the criteria for an exponential function that would make it purely increasing.
00:11:33.810 --> 00:11:44.580
What condition must there be on π§ for π of π₯ equals π§ over seven to the π₯ power, where π₯ is a positive number, to be an increasing function?
00:11:44.960 --> 00:11:54.490
For a function to be increasing, we know that as our values for π₯ increase, the output π of π₯ must itself also increase.
00:11:54.830 --> 00:12:06.820
And so how can we ensure that our function π of π₯ equals π§ over seven to the π₯ power is increasing over its entire domain, in other words, for all values of π₯?
00:12:07.080 --> 00:12:10.780
Well, letβs recall what we know about exponential functions.
00:12:11.070 --> 00:12:13.270
This is an exponential function.
00:12:13.580 --> 00:12:19.770
And the general form of an exponential function is π of π₯ equals π to the power of π₯.
00:12:20.050 --> 00:12:31.900
Now, as long as π is positive and a nonzero integer not equal to one, the function will be increasing if π is greater than one and decreasing if π is less than one.
00:12:32.210 --> 00:12:36.060
And so weβre going to let π be equal to π§ over seven.
00:12:36.250 --> 00:12:42.280
And then for our function to be increasing, π§ over seven must be greater than one.
00:12:42.640 --> 00:12:47.480
This is an inequality that we can solve just as we would solve any normal equation.
00:12:47.800 --> 00:12:50.300
Weβre going to multiply both sides by seven.
00:12:50.640 --> 00:12:55.190
π§ over seven times seven is π§, and one times seven is seven.
00:12:55.430 --> 00:13:04.630
And so π§ itself must be greater than seven for the function π of π₯ equals π§ over seven to the π₯ power to be an increasing function.
00:13:06.740 --> 00:13:09.070
Weβre now going to consider one final example.
00:13:09.380 --> 00:13:17.050
And weβre going to look to identify the increasing and decreasing intervals of a reciprocal function when weβve not been given the graph.
00:13:19.010 --> 00:13:26.830
Which of the following statements is true for the function β of π₯ equals negative one over seven minus π₯ minus five?
00:13:27.680 --> 00:13:34.960
Is it (A) β of π₯ is decreasing on the intervals negative β to seven and seven to β?
00:13:35.120 --> 00:13:42.030
Is it (B) β of π₯ is decreasing on the intervals negative β to negative seven and negative seven to β?
00:13:42.400 --> 00:13:49.280
(C) β of π₯ is increasing on the intervals negative β to negative seven and negative seven to β.
00:13:49.650 --> 00:13:56.060
Or (D) β of π₯ is increasing on the intervals negative β to seven and seven to β.
00:13:57.750 --> 00:14:01.590
If we look carefully, we see that β of π₯ is a reciprocal function.
00:14:01.690 --> 00:14:04.380
Itβs one over some polynomial.
00:14:04.790 --> 00:14:08.930
And so we know that there are probably going to be asymptotes on our graph.
00:14:09.270 --> 00:14:12.730
Letβs think about how we might sketch the graph of β of π₯.
00:14:12.970 --> 00:14:18.090
Weβll begin by starting with the function π of π₯ is equal to one over π₯.
00:14:18.380 --> 00:14:25.890
And then weβre going to consider the series of transformations that map the function one over π₯ onto the function β of π₯.
00:14:26.040 --> 00:14:28.160
Here is the function one over π₯.
00:14:28.410 --> 00:14:33.840
It has horizontal and vertical asymptotes made up of the π₯- and π¦-axis.
00:14:34.100 --> 00:14:39.710
Now weβll consider how we map π of π₯ onto one over negative π₯.
00:14:40.050 --> 00:14:43.450
This is represented by reflection in the π¦-axis.
00:14:43.700 --> 00:14:47.910
And then how do we map this onto the function one over seven minus π₯?
00:14:48.170 --> 00:14:57.160
Well, adding seven to the inner part of our composite function gives us a horizontal translation by negative seven.
00:14:57.210 --> 00:14:59.630
Thatβs a translation to the left seven units.
00:14:59.920 --> 00:15:05.350
Now, in doing this, our horizontal asymptote stays the same; itβs still the π₯-axis.
00:15:05.750 --> 00:15:09.190
But our vertical asymptote also shifts left seven units.
00:15:09.500 --> 00:15:15.840
And so it goes from being the π¦-axis, which is the line π₯ equals zero, to being the line π₯ equals negative seven.
00:15:16.230 --> 00:15:20.400
But of course, β of π₯ is negative one over seven minus π₯.
00:15:20.640 --> 00:15:23.940
This time, we reflect the graph in the π¦-axis.
00:15:24.160 --> 00:15:30.420
And so our horizontal asymptote remains unchanged, but our vertical asymptote is now at π₯ equals seven.
00:15:30.770 --> 00:15:35.480
Our final transformation maps this function onto β of π₯.
00:15:35.800 --> 00:15:39.350
Thatβs negative one over seven minus π₯ minus five.
00:15:39.640 --> 00:15:44.450
And now we move the entire graph we translated five units down.
00:15:44.980 --> 00:15:57.610
And so we now have the graph of β of π₯ of negative one over seven minus π₯ minus five, and weβre ready to decide whether the function is increasing or decreasing over the various intervals.
00:15:57.710 --> 00:16:04.840
Remember, if a function is decreasing, its graph will have a negative slope, and if itβs increasing, its graph will have a positive slope.
00:16:05.230 --> 00:16:15.620
As we move our values of π₯ from left to right, that is, from negative β all the way up to π₯ equals seven, we see that the graph is sloping downwards.
00:16:15.760 --> 00:16:19.020
It will approach negative β, but never quite reach it.
00:16:19.650 --> 00:16:26.960
Then as π₯ approaches positive β from seven, the graph continues to slope downwards.
00:16:27.240 --> 00:16:30.080
This time, though, it approaches negative five.
00:16:30.370 --> 00:16:37.960
And so the function is definitely decreasing over these intervals from negative β to seven and seven to β.
00:16:38.380 --> 00:16:47.140
Since the function itself cannot take a value of π₯ equals seven, and this is why we have the horizontal asymptote, then we want to include open intervals.
00:16:47.140 --> 00:16:48.540
Those are the round brackets.
00:16:48.880 --> 00:16:58.260
And so the correct answer must be (A), β of π₯ is decreasing on the open interval negative β to seven and seven to β.
00:17:00.120 --> 00:17:02.440
Weβll now recap the key points from this lesson.
00:17:03.660 --> 00:17:10.710
In this video, we learned that a function is increasing if π of π₯ increases as the value of π₯ increases.
00:17:11.000 --> 00:17:16.260
Over these intervals, the graph of the function will have a positive slope or a positive gradient.
00:17:16.520 --> 00:17:24.340
And then, if the function decreases as π₯ increases, we say itβs decreasing and the graph will have negative slope.
00:17:24.510 --> 00:17:30.980
Remember, we can say that a function is strictly increasing or strictly decreasing, if there are no flat bits on the graph at all.
00:17:31.290 --> 00:17:42.660
Finally, we saw that a function is constant if the value of π of π₯ remains unchanged as the value of π₯ increases and the graph of a constant function looks like a horizontal line.