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Composition of Functions:
Composing with Sets of Points (page 1 of 6)
Composing with Sets of Points (page 1 of 6)
Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition
Until now, given a function f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.
In function composition, you're plugging entire functions in for the x. In other words, you're always getting 'fancy'. But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:
Login to access your 1&1 e-mail account and read your e-mail online with 1&1 Webmail. Subtraction is removing some objects from a group. The meaning of 5-3=2 is that three objects are taken away from a group of five objects and two objects remain. The subtraction facts for 0, 1, and 2 are.
- Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and
let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.
Find (i)f (1), (ii) g(–1), and (iii) (gof )(1).
(i) This type of exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:
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f (1) = –1
(ii) The point in the g(x) set of point with x = –1 is the point (–1, –2), so:
g(–1) = –2
(iii) What is '(gof )(1)'? This is read as 'g-compose-f of 1', and means 'plug 1 into f, evaluate, and then plug the result into g'. The computation can feel a lot easier if I use the following, more intuitive, formatting:
(gof )(1) = g( f(1))
Now I'll work in steps, keeping in mind that, while I may be used to doing things from the left to the right (because that's how we read), composition works from the right to the left (or, if you prefer, from the inside out). So I'll start with the x = 1. I am plugging this into f(x), so I look in the set of f(x) points for a point with x = 1. The point is (1, –1). This tells me that f(1) = –1, so now I have: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
(gof )(1) = g( f(1)) = g(–1)
Working from the right back toward the left, I am now plugging x = –1 (from 'f(1) = –1') into g(x), so I look in the set of g(x) points for a point with x = –1. That point is (–1, –2). This tells me that g(–1) = –2, so now I have my answer:
(gof )(1) = g( f(1)) = g(–1) = –2
Note that they never told us what were the formulas, if any, for f(x) or g(x); we were only given a list of points. But this list was sufficient for answering the question, as long as we keep track of our x- and y-values.
- Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and
let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.
Find (i) ( fog)(0), (ii)( fog)(–1), and (iii)(gof )(–1).
(i) To find ( fog)(0), ('f-compose-g of zero'), I'll rewrite the expression as:
( fog)(0) = f(g(0))
This tells me that I'm going to plug zero into g(x), simplify, and then plug the result into f(x). Looking at the list of g(x) points, I find (0, 2), so g(0) = 2, and I need now to find f(2). Looking at the list of f(x) points, I find (2, –3), so f(2) = –3. Then:
( fog)(0) = f(g(0)) = f(2) = –3
(ii) The second part works the same way:
( fog)(–1) = f(g(–1)) = f(–2) = 3
(iii) I can rewrite the composition as (gof )(–1) = g( f(–1)) = g(1).
Uh-oh; there is no g(x) point with x = 1, so it is nonsense to try to find the value of g(1). In math-speak, g(1) is 'not defined'; that is, it is nonsense.Then (gof )(–1) is also nonsense, so the answer is:
(gof )(–1) is undefined.
Part (iii) of the above example points out an important consideration regarding domains and ranges. It may be that your composed function (the result you get after composing two other functions) will have a restricted domain, or at least a domain that is more restricted than you might otherwise have expected. This will be more important when we deal with composing functions symbolically later.
Another exercise of this type gives you two graphs, rather than two sets of points, and has you read the points (the function values) from these graphs.
- Given f(x) and g(x) as shown below, find ( fog)(–1).
In this case, I will read the points from the graph. I've been asked to find ( fog)(–1) = f(g(–1)). This means that I first need to find g(–1). So I look on the graph of g(x), and find x = –1. Tracing up from x = –1 to the graph of g(x), I arrive at y = 3. Then the point (–1, 3) is on the graph of g(x), and g(–1) = 3.
Now I plug this value, x = 3, into f(x). To do this, I look at the graph of f(x) and find x = 3. Tracing up from x = 3 to the graph of f(x), I arrive at y = 3. Then the point (3, 3) is on the graph of f(x), and f(3) = 3.
Then( fog)(–1) = f(g(–1)) = f(3) = 3.
- Given f(x) and g(x) as shown in the graphs below, find (gof )(x) for integral values of x on the interval –3 <x< 3.
f(x): | g(x): |
This is asking me for all the values of (gof )(x) = g( f(x)) for x = –3, –2, –1, 0, 1, 2, and 3. So I'll just follow the points on the graphs and compute all the values:
(gof )(–3) = g( f(–3)) = g(1) = –1
I got this answer by looking at x = –3 on the f(x) graph, finding the corresponding y-value of 1 on the f(x) graph, and using this answer as my new x-value on the g(x) graph. That is, I looked at x = –3 on the f(x) graph, found that this led to y = 1, went to x = 1 on the g(x) graph, and found that this led to y = –1. Similarly:
(gof )(–2) = g( f(–2)) = g(–1) = 3
(gof )(–1) = g( f(–1)) = g(–3) = –2
(gof )(0) = g( f(0)) = g(–2) = 0
(gof )(1) = g( f(1)) = g(0) = 2
(gof )(2) = g( f(2)) = g(2) = –3
(gof )(3) = g( f(3)) = g(3) = 1
(gof )(–1) = g( f(–1)) = g(–3) = –2
(gof )(0) = g( f(0)) = g(–2) = 0
(gof )(1) = g( f(1)) = g(0) = 2
(gof )(2) = g( f(2)) = g(2) = –3
(gof )(3) = g( f(3)) = g(3) = 1
You aren't generally given functions as sets of points or as graphs, however. Generally, you have formulas for your functions. So let's see what composition looks like in that case..
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Cite this article as: | Stapel, Elizabeth. 'Composing with Sets of Points.' Purplemath. Available from https://www.purplemath.com/modules/fcncomp.htm. Accessed [Date] [Month] 2016 |
A circle is easy to make:
Draw a curve that is 'radius' away
from a central point.
from a central point.
And so:
All points are the same distance
from the center.
from the center.
In fact the definition of a circle is
Circle: The set of all points on a plane that are a fixed distance from a center.
Circle on a Graph
Let us put a circle of radius 5 on a graph:
Now let's work out exactly where all the points are.
We make a right-angled triangle:
And then use Pythagoras:
x2 + y2 = 52
There are an infinite number of those points, here are some examples:
x | y | x2 + y2 |
---|---|---|
5 | 0 | 52 + 02 = 25 + 0 = 25 |
3 | 4 | 32 + 42 = 9 + 16 = 25 |
0 | 5 | 02 + 52 = 0 + 25 = 25 |
−4 | −3 | (−4)2 + (−3)2 = 16 + 9 = 25 |
0 | −5 | 02 + (−5)2 = 0 + 25 = 25 |
In all cases a point on the circle follows the rule x2 + y2 = radius2
We can use that idea to find a missing value
Example: x value of 2, and a radius of 5
Values we know:22 + y2 = 52
Simplescrobbler 2 0 18
Square root both sides: y = ±√(52 − 22)
y ≈ ±4.58..
(The ± means there are two possible values: one with + the other with −)
And here are the two points:
More General Case
Now let us put the center at (a,b)
So the circle is all the points (x,y) that are 'r' away from the center (a,b).
Now lets work out where the points are (using a right-angled triangle and Pythagoras):
Simplescrobbler 2 0 1 9
It is the same idea as before, but we need to subtract a and b:
(x−a)2 + (y−b)2 = r2
And that is the 'Standard Form' for the equation of a circle!
It shows all the important information at a glance: the center (a,b) and the radius r.
Example: A circle with center at (3,4) and a radius of 6:
Start with:
(x−a)2 + (y−b)2 = r2
Put in (a,b) and r:
(x−3)2 + (y−4)2 = 62
We can then use our algebra skills to simplify and rearrange that equation, depending on what we need it for.
Try it Yourself
'General Form'
But you may see a circle equation and not know it!
Because it may not be in the neat 'Standard Form' above.
As an example, let us put some values to a, b and r and then expand it Dvdfab 10 0 all in one for mac free download.
Example: a=1, b=2, r=3:(x−1)2 + (y−2)2 = 32
Gather like terms:x2 + y2 − 2x − 4y + 1 + 4 − 9 = 0
And we end up with this:
x2 + y2 − 2x − 4y − 4 = 0
It is a circle equation, but 'in disguise'!
So when you see something like that think 'hmm .. that might be a circle!'
In fact we can write it in 'General Form' by putting constants instead of the numbers:
Note: General Form always has x2 + y2 for the first two terms.
Going From General Form to Standard Form
Now imagine we have an equation in General Form:
x2 + y2 + Ax + By + C = 0
How can we get it into Standard Form like this?
(x−a)2 + (y−b)2 = r2
The answer is to Complete the Square (read about that) twice .. once for x and once for y:
Example: x2 + y2 − 2x − 4y − 4 = 0
Put xs and ys together:(x2 − 2x) + (y2 − 4y) − 4 = 0
Now complete the square for x (take half of the −2, square it, and add to both sides):
(x2 − 2x + (−1)2) + (y2 − 4y) = 4 + (−1)2
And complete the square for y (take half of the −4, square it, and add to both sides):
(x2 − 2x + (−1)2) + (y2 − 4y + (−2)2) = 4 + (−1)2 + (−2)2
Tidy up:
Finally:(x − 1)2 + (y − 2)2 = 32
And we have it in Standard Form!
Simplescrobbler 2 0 1 Download
(Note: this used the a=1, b=2, r=3 example from before, so we got it right!)
Unit Circle
If we place the circle center at (0,0) and set the radius to 1 we get:
(x−a)2 + (y−b)2 = r2 (x−0)2 + (y−0)2 = 12 x2 + y2 = 1 Which is the equation of the Unit Circle |
How to Plot a Circle by Hand
1. Plot the center (a,b)
2. Plot 4 points 'radius' away from the center in the up, down, left and right direction
3. Sketch it in!
Example: Plot (x−4)2 + (y−2)2 = 25
The formula for a circle is (x−a)2 + (y−b)2 = r2
So the center is at (4,2)
And r2 is 25, so the radius is √25 = 5
So we can plot:
- The Center: (4,2)
- Up: (4,2+5) = (4,7)
- Down: (4,2−5) = (4,−3)
- Left: (4−5,2) = (−1,2)
- Right: (4+5,2) = (9,2)
Now, just sketch in the circle the best we can!
How to Plot a Circle on the Computer
We need to rearrange the formula so we get 'y='.
We should end up with two equations (top and bottom of circle) that can then be plotted.
![Simplescrobbler Simplescrobbler](https://image.winudf.com/v2/image1/Y29tLmFkYW0uYXNsZm1zX3NjcmVlbl83XzE1Njc5MTAwODdfMDk2/screen-7.jpg?fakeurl=1&type=.jpg)
Example: Plot (x−4)2 + (y−2)2 = 25
So the center is at (4,2), and the radius is √25 = 5
Rearrange to get 'y=':
Move (x−4)2 to the right: (y−2)2 = 25 − (x−4)2
(notice the ± 'plus/minus' ..
there can be two square roots!)
there can be two square roots!)
So when we plot these two equations we should have a circle:
- y = 2 + √[25 − (x−4)2]
- y = 2 − √[25 − (x−4)2]
Try plotting those functions on the Function Grapher.
It is also possible to use the Equation Grapher to do it all in one go.