GRAPHS OF FUNCTIONS
Some Functions and their Inverses
An invertible function is a function that can be inverted. An invertible function must satisfy the condition that each element in the domain corresponds to one distinct element that no other element in the domain corresponds to. That is, all of the elements in the domain and range are paired-up in monogomous relationships - each element in the domain pairs to only one element in the range and each element in the range pairs to only one element in the domain. Thus, the inverse of a function is a function that looks at this relationship from the other viewpoint. So, for all elements a in the domain of f(x), the inverse of f(x) (notation: f-1(x)) satisfies:
f(a)=b implies f-1(b)=a
And, if you do the slightest bit of manipulation, you find that:
Yielding the identity function for all inputs in the domain.
When we graph functions and their inverses, we find that they mirror along the line x=y. This is only logical. From our definition, we know that for each (a,b) in f(x) there will be a (b,a) in f-1(x):
As an exercise, draw a graph of any one-to-one function you know - or make one up. Then, take some points
and plot them at
in a different color. This second curve is the inverse of the first.
Lines can be easily manipulated to find inverses. Except for horizontal and vertical lines, each point x on a line corresponds to one y. I will illustrate the general procedure for finding the inverse of a function with a line. Let us take the line described by y = 2x-5.
General procedure for finding the inverse of a function:
- Interchange the variables - First, we will exchange the variables. We do this because we want to find the function that goes the other way, by mapping the old range onto the old domain. So our new equation is x=2y-5.
- Solve for y -The rest is simply solving for the new y, which gives us:
2y-5 = x
2y = x+5
y = (x+5)/2
Hence, y-1(x) = (x+5)/2
Now that that is out of the way, we have bigger inversion problems to worry about. What happens if we try to find the inverse of a parabola? Well, look at the graph:
What happens is that, because a parabola is not a one-to-one the inverse can't exist because for various values of
has to take on
values! To solve this problem in taking inverses, in many cases, people decide to simply limit the domain. For instance, by limiting the domain of the parabola
to values of
, we can say that the function's inverse is
means the square root of x or
) This is done to let the trigonometric functions have inverses.
As you can see, we can't take the inverse of sin(x) because it is not a one-to-one function. However, we can take the inverse of a subset of
with the domain of
-/2 to /2
. The new function inverse we get is called
Even and Odd Functions
Even functions are functions for which the left half of the plane looks like the mirror image of the right half of the plane. Odd functions are functions where the left half of the plane looks like the mirror image of the right half of the plane, only upside-down.
Mathematically, we say that a function f(x) is even if f(x)=f(-x) and is odd if f(-x)=-f(x).
|Some even functions||Some odd functions|
Sine(x) - the most common periodic function with period 2
Tangent(x) - another periodic function with period
Periodic functions are functions that repeat over and over, or cycle on a specific period. This is expressed mathematically that
A function is periodic if "there exists some number p>0 such that f(x)=f(x+p) for all possible values of x" [1.7,p.112]
The fundamental period of a function is the length of a smallest continuous portion of the domain over which the function completes a cycle. That is, it's the smallest length of domain that if you took the function over that length and made an infinite number of copies of it, and laid them end to end, you would have the original function.
For example, let's say that we have some imaginary function f(x) and that f(0.1)=2.35 if f(x) is a periodic function with period 2, then f(2.1)=2.35, because 2.1 is exactly 0.1+2. That is, after 2 units, you're back where you started. This continues up and down the number line because the function is periodic, so:
We can generalize this by saying that for our function f(x), f(2*k+0.1)=2.35, k integer. To further generalize the pattern, we can define an arbitrary periodic function g(x) with period p by saying that:
f(p*k+a) = f(a) is true for all a real, k integer.
A property of some periodic funtions that cycle within some definite range is that they have an amplitude in addition to a period. The amplitude of a periodic function is the distance between the highest point and the lowest point, divided by two. For example, sin(x) and cos(x) have amplitudes of 1.
Combining Periodic Functions
Because sine and cosine are both periodic and have the same period, when you add them up, subtract them, multiply them, etc you get functions that are also periodic. This can be proven with some care - PROOF - Let us assume that:
f(p*k+a) = f(a) is true for all a real, k integer.
Simply multiplying each side by some constant does not change the equation, nor does adding or subtracting some constant to each side change the periodicity:
K = f(p*k+a) = f(a)
c*K = c*K
K + c = K + c
But this is simple equation manipulation. What about when we mix different functions? If we have two functions,
f(x) & g(x)
, with the same period, we can throw them together any way we want because, if their period is p, and at any value
|Homework Assignment: Extend this proof for multiplication - use h(x) = f(x)*g(x)||f(a) = c1; |
g(a) = c2, so
f(a) + g(a) = c1 + c2 = C
f(a) + g(a) = C
Also, because they are periodic, for k integer:
f(a) = f(p*k+a) = c1
g(a) = g(p*k+a) = c2
f(p*k+a) + g(p*k+a) = c1 + c2 = C
To make things clearer, many people use an auxilury function h(x):
h(x) = f(x) + g(x)
h(a) = f(x) + g(x) = C
h(p*k+a) = f(p*k+a) + g(p*k+a) = C
Thus, h(a) = h(p*k+a)
What about functions that don't have the same fundamental period? Well, that's a different story. Let's say we have two (new) periodic functions
f(x) and g(x)
, with periods
p and r
f(p*k+a) = f(a) is true for all a real, k integer.
g(r*k+a) = g(a) is true for all a real, k integer.
Now we can't construct that nice
like we did because we have clashing periods. (Not to be mistaken with cat fights resulting from two women with PMS at the same time) So what can we do? Functions don't have just one period! If a function repeats every 2 units, then it will also repeat every 6 units, won't it? So if we have one function with fundamental period 2, and another with fundamental period 8, we've got no problem because 8 is a multiple of 2, and both functions will cycle every 8 units.
So, if we can "patch up" the periods to be the same, we know that if we combine them, we'll get a function with the "patched up" period.
What about if
has period 3 and
has period 4? We can easily see that in 12 units, both will cycle, so they're fine. What about fractions? Let's say that we had functions with periods
13/12 - f(x)
2/21 - g(x)
. What we need are two numbers of periods that we can multiply by the periods to get some common, "patched-up period". First, we can simplify the problem by multiplying each period by its denominator to find whole number periods. So we know that
has a period of 13 (in 12 fundamental periods) and
a period of 2 (in 21 fundamental periods). Now we can simply do what we did in the 3 and 4 case and multiply by both periods to find a period for the new combination function. So
has a period of 26. There's a walk through pseudo-proof for the following Theorem (generalize the proof for homework!)
If two periodic functions, f(x) and g(x), have rational periods, then any addition or multiplication combination of the functions (not composition!) will also be periodic.
Now composition of different irrational numbers, like e and , is a different story. Someone from number theory - a proof would be greatly appreciated.
Also, for those who like digging, I challenge you to prove or disprove that if f(x) is a periodic function and g(x) is not a periodic function, then g(f(x)) is periodic and f(g(x)) is not.
|y = sqrt(x)||y = log2(x)|
The graph of the square root starts at the point (0, 0) and then goes off to the right. On the other hand, the graph of the log passes through (1, 0), going off to the right but also sliding down the positive side of the y-axis. Remembering that logs are the inverses of exponentials, this shape for the log graph makes perfect sense: the graph of the log, being the inverse of the exponential, would just be the "flip" of the graph of the exponential:
|y = 2x||y = log2(x)|
|comparison of the two graphs, |
showing the inversion line in red
It is fairly simple to graph exponentials. For instance, to graph y = 2x, you would just plug in some values for x, compute the corresponding y-values, and plot the points. But how do you graph logs? There are two options. Here is the first: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
- Graph y = log2(x).
In order to graph this "by hand", I need first to remember that logs are not defined for negative x or for x = 0. Because of this restriction on the domain (the input values) of the log, I won't even bother trying to find y-values for, say, x = –3 or x = 0. Instead, I'll start with x = 1, and work from there, using the definition of the log.
- Since 20 = 1, then log2(1) = 0, and (1, 0) is on the graph.
- Since 21 = 2, then log2(2) = 1, and (2, 1) is on the graph.
- Since 3 is not a power of 2, then log2(3) will be some messy value. So I won't bother with graphing x = 3.
- Since 22 = 4, then log2(4) = 2, and (4, 2) is on the graph.
- Since 5, 6, and 7 aren't powers of 2 either, I'll skip them and move up to x = 8.
- Since 23 = 8, then log2(8) = 3, so (8, 3) is on the graph.
- The next power of 2 is 16: since 24 = 16, then log2(16) = 4, and (16, 4) is on the graph.
- The next power of 2, x = 32, is too big for my taste; I don't feel like drawing my graph that wide, so I'll quit at x = 16.
The above gives me the point (1, 0) and some points to the right, but what do I do for x-values between 0 and 1? For this interval, I need to think in terms of negative powers and reciprocals. Just as the left-hand "half" of the exponential function had few graphable points (the rest of them being too close to the x-axis), so also the bottom "half" of the log function has few graphable points, the rest of them being too close to the y-axis. But I can find a few:
Since 2–1 = 1/2 = 0.5, then log2(0.5) = –1, and (0.5, –1) is on the graph.
Since 2–2 = 1/4 = 0.25, then log2(0.25) = –2, and (0.25, –2) is on the graph.
Since 2–3 = 1/8 = 0.125, then log2(0.125) = –3, and (0.125, –3) is on the graph.
The next power of 2 (as x moves in this direction) is 1/16 = 2–4, but the x-value for the point (0.0625, –4) seems too small to bother with, so I'll quit with the points I've already found.
Listing these points gives me my T-chart:
Drawing my dots and then sketching in the line (remembering not to go to the left of the y-axis!), I get this graph: