WEBVTT
00:00:02.650 --> 00:00:10.090
Find the domain and range of the function π of π₯ is equal to the absolute value of three π₯ minus one plus three π₯ plus five.
00:00:10.560 --> 00:00:15.920
In this question, weβre given a function π of π₯ and weβre asked to determine the domain and the range of this function.
00:00:16.090 --> 00:00:20.810
So to answer this question, letβs start by recalling what we mean by the domain and range of a function.
00:00:21.010 --> 00:00:25.450
First, the domain of a function is the set of all input values for that function.
00:00:25.480 --> 00:00:30.520
Second, the range of a function is the set of all output values for the function given its domain.
00:00:30.820 --> 00:00:34.460
And thereβs many different ways of determining the domain and range of a function.
00:00:34.530 --> 00:00:36.910
One way of doing this is to sketch its graph.
00:00:37.140 --> 00:00:44.380
To help us sketch the graph of π¦ is equal to π of π₯, letβs call the function the absolute value of three π₯ minus one π of π₯.
00:00:44.640 --> 00:00:49.310
Then, since this contains an absolute value, we can rewrite this as a piecewise function.
00:00:49.480 --> 00:00:57.600
To do this, we recall the absolute value of a function changes definition depending on whether itβs taking the absolute value of a positive or a negative number.
00:00:57.920 --> 00:01:04.640
Therefore, if three π₯ minus one is greater than or equal to zero, the absolute value of three π₯ minus one is just three π₯ minus one.
00:01:04.880 --> 00:01:11.170
However, if three π₯ minus one is less than zero, when we take its absolute value, we need to multiply this by negative one.
00:01:11.300 --> 00:01:15.280
And negative one times three π₯ minus one is one minus three π₯.
00:01:15.460 --> 00:01:18.940
This then gives us a piecewise definition of the function π of π₯.
00:01:19.060 --> 00:01:23.450
And itβs worth noting we can simplify this definition by simplifying its subdomains.
00:01:23.590 --> 00:01:27.580
We can add one to both inequalities and then divide both inequalities through by three.
00:01:27.580 --> 00:01:33.890
We get π₯ is greater than or equal to one-third and π₯ is less than one-third, giving us the following expression for π of π₯.
00:01:33.920 --> 00:01:38.050
Now, we can use this to find a piecewise definition of the function π of π₯.
00:01:38.120 --> 00:01:43.010
First, we see that our function π of π₯ is equal to π of π₯ plus three π₯ plus five.
00:01:43.150 --> 00:01:48.220
This means the output of our function π of π₯ is going to depend on the output of our function π of π₯.
00:01:48.330 --> 00:01:53.130
If π₯ is greater than or equal to one-third, π of π₯ is going to output three π₯ minus one.
00:01:53.200 --> 00:01:56.280
We can then substitute this into our function π of π₯.
00:01:56.310 --> 00:02:04.640
If π₯ is greater than or equal to one-third, π of π₯ is equal to three π₯ minus one plus three π₯ plus five, which we can simplify as six π₯ plus four.
00:02:04.810 --> 00:02:09.740
However, if our input value of π₯ is less than one-third, then π of π₯ is one minus three π₯.
00:02:09.870 --> 00:02:12.590
We can then substitute this into our function π of π₯.
00:02:12.590 --> 00:02:19.130
If π₯ is less than one-third, π of π₯ is equal to one minus three π₯ plus three π₯ plus five, which we can simplify.
00:02:19.180 --> 00:02:24.640
Negative three π₯ plus three π₯ is zero, so π of π₯ is just equal to six when π₯ is less than one-third.
00:02:24.830 --> 00:02:35.550
And if π of π₯ is equal to the constant value of six when π₯ is less than one-third and π of π₯ is equal to six π₯ plus four when π₯ is greater than or equal to one-third, π of π₯ is a piecewise-defined function.
00:02:35.710 --> 00:02:40.280
And now we can sketch the graph π¦ is equal to π of π₯ to determine its domain and range.
00:02:40.440 --> 00:02:46.390
However, before we do this, recall that the domain of a piecewise-defined function is the union of its subdomains.
00:02:46.490 --> 00:02:51.140
So in fact, we can determine the domain of our function π of π₯ from its piecewise definition.
00:02:51.170 --> 00:02:55.610
Itβs all values of π₯ less than one-third or values of π₯ greater than or equal to one-third.
00:02:55.910 --> 00:02:58.140
In other words, itβs all real values of π₯.
00:02:58.310 --> 00:03:00.600
Now we need to determine the range of this function.
00:03:00.630 --> 00:03:02.280
Weβll do this by sketching its graph.
00:03:02.570 --> 00:03:06.860
First, the function is a constant value of six when π₯ is less than one-third.
00:03:07.210 --> 00:03:15.100
And since the output values are a constant value of six, this is the horizontal line π¦ is equal to six, where our input values of π₯ must be less than one-third.
00:03:15.390 --> 00:03:18.470
And we represent this by a hollow circle when π₯ is one-third.
00:03:18.760 --> 00:03:24.660
Next, when π₯ is greater than or equal to one-third, our function π of π₯ is a linear function: six π₯ plus four.
00:03:24.790 --> 00:03:29.660
And to sketch this linear function, letβs start by finding its endpoint, the value when π₯ is one-third.
00:03:29.820 --> 00:03:34.880
We substitute π₯ is equal to one-third into the linear function to get six times one-third plus four.
00:03:34.910 --> 00:03:37.840
And if we evaluate this, we see itβs equal to six.
00:03:38.080 --> 00:03:43.000
Therefore, the endpoint of this linear function is the point with coordinates one-third, six.
00:03:43.190 --> 00:03:49.640
And this is the hollow dot we already had on our graph, which means we can now fill this in because this is now a point on the graph of our function.
00:03:49.640 --> 00:03:51.740
Itβs the endpoint of the linear portion.
00:03:52.020 --> 00:03:54.550
We could now sketch the rest of this graph accurately.
00:03:54.580 --> 00:03:57.670
However, weβll see itβs not necessary to determine its range.
00:03:57.800 --> 00:04:00.470
All we need to know is that this line has positive slope.
00:04:00.470 --> 00:04:01.770
Its slope is six.
00:04:01.890 --> 00:04:05.720
And so as π₯ approaches β, the line is going to approach β.
00:04:05.890 --> 00:04:08.830
Weβre now ready to determine the range of this function from its graph.
00:04:08.880 --> 00:04:13.190
Remember, the range of the function is the set of all possible output values of the function.
00:04:13.410 --> 00:04:21.810
And when we sketch the graph of a function, the π₯-coordinate of the point on the curve tells us the input value and the π¦-coordinate tells us the corresponding output value.
00:04:22.140 --> 00:04:26.190
Therefore, the π¦-coordinates are points on the curve that are telling us the range of our function.
00:04:26.400 --> 00:04:30.290
For example, from the graph, we can see that six is an output value of our function.
00:04:30.410 --> 00:04:32.020
In fact, itβs the lowest output.
00:04:32.020 --> 00:04:34.440
Itβs the lowest π¦-coordinate of a point on the graph.
00:04:34.640 --> 00:04:37.560
And we also know that our line extends to β.
00:04:37.610 --> 00:04:42.050
Therefore, any value greater than or equal to six is a possible output of the function.
00:04:42.190 --> 00:04:45.780
So the range of this function contains all values greater than or equal to six.
00:04:45.810 --> 00:04:47.460
Remember, we need to write this as a set.
00:04:47.460 --> 00:04:50.990
This is the left-closed, right-open interval from six to β.
00:04:51.230 --> 00:05:00.930
Therefore, we were able to show the domain of the function π of π₯ given to us in the question is the set of real numbers, and its range is the left-closed, right-open interval from six to β.