7.3 Modeling a function
7.3.1 Inverse Functions
Let is recall the various terms used in connection with a function.
Suppose that we have a function f : A → B where A, B are certain sets of numbers (real or complex.)
We call A the domain of the function and call B its target. The range of the function is defined as the set
{y ∈ B | y = f(x) for some x ∈ A}.
In general, determining the range of a function involves extra calculations.
We say that the function f is “one to one” if different values of x give different values of f (x).
This can be reworded as
One to one property. If f (x1) = f (x2) then, x1 = x2
We say that the function f is onto B if B is the same as the range. The set B is not mentioned, if clear from the context.
Explicitly, this means:
Onto property. For every y ∈ B there is some x ∈ A such that f(x) = y A function g : B → A is said to be the inverse of the function f if it satisfies: Inverse conditions. f (g(y)) = y for all y ∈ B and g(f(x)) = x for all x ∈ A
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Finally, we have the test for existence of an inverse function.
A function f : A → B has an inverse if and only if f is both one to one and onto.
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If a graph of a real y = f (x) available, then there is a graphical way of deciding its “one to one” and “onto” properties.
If we assume that both the target and the domain of the function are the same ℜ-the set of all real numbers, then the simple test is this:
3Why do we use two different letters x, y in this definition? It is meant as a help in calculation.
We are thinking of the function as y = f (x), so elements of the domain are denoted by x and elements of the target are denoted by y. Often both the domain and the target are the same, but this convention keeps us thinking of the correct set.
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• The function is onto a set A ⊂ ℜ if and only for every a ∈ A, the horizontal line y = a meets the graph at least once.
• The function is one to one if and only for every b ∈ ℜ, the horizontal line y = b meets the graph at most once.
Here is the modification of the test for more special targets and domains.
If the domain is a set D smaller than ℜ, then in the “ one to one ” test, we require a horizontal line to meet the graph at most once over the domain D.
Similarly, if the target is smaller than ℜ then, the horizontal lines y = b need to be chosen only with b in the target.
Special notation for inverse function.
Suppose f : A → B is a given function which has an inverse, say g. Then it is customary to denote g by the symbol f−1. Thus, we have
f (f−1(y)) = y and f−1(f (x)) = x.
This notation can lead to confusion and one needs to be careful with it. Thus f−1(x)
is not the same as f (x)−1 = 1
f (x). Indeed, it is usually very different from it. We will avoid this notation, unless dictated by tradition.
Let us illustrate the above concepts by a few examples.
1. Analyzing a real function. Consider f : ℜ → ℜ defined by f(x) = x2.
• Determine the range of f.
• Determine if the function f is one to one. • Determine the inverse of f, if it exists. Answer.
Set y = f (x) = x2. We want to find those values of y for which we can solve
the equation y = x2. We already know that the solution of this equation is:
x = ±√y if y ≥ 0 and no real solution if y < 0.
• Suppose that f has an inverse g. If f(x) = y, then applying g to both sides, we see that g(f (x)) = g(y) and by definition of the inverse function, we get x = g(y).
Now, if f (x1) = f (x2) = y, then we see that x1= g(y) = x2, and this proves f is one to one.
Also, given any y ∈ B we see by definition that y = f(g(y)) so y is the image of g(y) by f. This proves that f is onto B.
• Conversely, if f is onto B, we see that for every y ∈ B, there is at least one x ∈ A such that y = f (x). Moreover, by one to one property, there is a single x such that y = f (x). We define g(y) = x and verify the two conditions easily.
126 CHAPTER 7. FUNCTIONS
Thus the range is
ℜ+= {y ∈ ℜ | y ≥ 0} = [0, ∞).
Since both x = 1 and x = −1 lead to y = f(x) = 1, we see that the function is not one to one.
By the known test for inverse functions, the inverse function does not exist! 2. Analyzing a complex function. Consider f : C → C defined by
f (x) = x2.
• Determine the range of f.
• Determine if the function f is one to one. • Determine the inverse of f, if it exists.
Answer. When working with complex numbers, we see that the equation y = x2 always has solutions for any y.
Indeed, if y = 0 then take x = 0. If y 6= 0, then write y = r exp(iθ) using Euler’s representation, where r = |y| > 0 and θ is some real number which is its argument Arg(y).
Then it is easy to show that x1 = r exp(iθ/2) and x2 = r exp(i(θ + π)/2)
satisfy f (x1) = f (x2) = y. Thus the function is onto but not one to one.
In particular it does not have an inverse.
3. Changing the domain or the target. We remark that a function can be made “onto” by trimming its target down to its range. We also note that a function can be made “one to one” by trimming down its domain, but this process can be rather complicated and may lead to unwieldy domains. Here are a few examples.
The same real function y = x2 is onto if we set our target as ℜ
+ the set of non
negative reals. If we consider the same formula but take our domain and target as both ℜ+, then it is one to one and onto. Its inverse is given by
g(x) =√x which is defined on the domain ℜ+ and by √x, we mean the non
negative square root.
4. Finding the inverse. In general, for a complicated function, verifying the conditions for an inverse is difficult. You will see some of the hard calculations in the Appendix.
There is a simple procedure which may sometimes succeed and we describe it next.
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Given a function f (x), try to solve the equation y = f (x) for x. The range would be those values of y for which the equation has a solution and the function would be one to one if the solution is unique.
Thus, the solution, written as x = g(y) would give the necessary inverse function, namely g(x) (i.e. the function obtained by replacing y by x in g(y)). Here is an example of a function f (x) for which this simple idea works.
Consider f (x) = 3x + 5. Set y = 3x + 5 and solve for x. This gives x = y − 53 . This led to a well defined answer, so the inverse function exists. It is defined so that g(y) = x, i.e. g(y) = y − 53 .
Now we change the variable to x, and write g(x) = x − 53 .
In general, if y = f (x) = ax + b where a 6= 0, then the the inverse function is given by x = y − ba . Thus g(y) = y − ba or g(x) = x − ba .
5. Inverse functions on a calculator. Many routine functions appear to have inverse keys on a calculator. The existence of the keys does not mean the functions have an inverse. It only means that under a suitably chosen domain and target they have an inverse known to the calculator.
The simplest example of this is the square root function.
Recall that the square root function is defined as the inverse of the square function which may be defined as
f := ℜ → ℜ where f(x) = x2.
It is known that the range is the set of non negative reals and the function is not one to one. (For example, f (2) = f (−2) = 4.)
The calculator will declare an error if you ask for √x when x is outside the range , i.e. x < 0.
It will, however, give √4 = 2, not bothering to remind you that −2 is also a possible number whose square is 4. Thus, it is acting as if the domain of the function f is also the set of non negative reals.
It is your task to think about which solution of f (x) = 4 is needed. This problem becomes more acute when trigonometric functions are used, since they have many values of x leading to the same y.