Given that we are going to talk about inverse functions (sometimes called reciprocals), we should first make sure that the concept of a function is clear. Although we have already worked with functions several times in both Maths and Physics, have we asked ourselves at any point what a function actually is? Could you give an accurate definition of a function? We all have an intuitive idea of what a function is, but we need to try and express it in mathematical terms. It may seem as if we are leaving our intuition aside by thinking of a formal definition. However, we should try to see how our intuition links to the formal definition, which has already been expressed in straightforward language by those who thought out and/or sensed the definition intuitively.

Throughout history we have seen several definitions of a function. In this unit we are going to refer to the most widely used definition in the field of Modern Mathematics.

Before giving the definition of the function we should remember that:

• RxR = R², Cartesian product of R by R, is the set of all ordered pairs (x, y), where both x and y are real numbers. In other words:

  RxR = R2 = { (x, y) / x, y Î R }

• An ordered pair of numbers can be represented on the Coordinate System (two straight lines which cross each other at a right angle). We call the first element in a pair of ordered numbers the first coordinate or x-coordinate and the second is the second coordinate or y-coordinate.

• A set can be defined by giving a list of all its members (listing all the elements individually) or by describing a certain property which needs to be satisfied (defining through understanding).

EXAMPLES:

• f = { (1, 2), (2, 4), (3, -1), (4, 2) } is a function.
• g = { (1, 2), (2, 1), (1, 3), (3, -1) } is not a function as the pairs of numbers (1, 2) and (1, 3) have the same first coordinate and according to the definition 2 is equal to 3 (2=3), which is not true.

When a set is written out as a list of pairs of numbers it is easy to see whether we are dealing with a function or not. However, if the function is defined by a property it can be very difficult to determine whether it is a function or not. The ability to distinguish will depend on your mathematical knowledge.

EXAMPLES:

1. f = { (x, y) / y=2x } is a function, as the value of y can only be determined from x.
2. g = { (a, b) / a²+b²=9 } is not a function as the pairs of numbers (0, 3) and (0, -3) have the same first coordinate and a different second coordinate.
3. h = { (a, b) / a²+b³=16 } is a function; by getting b on its own we can see that each value of a determines one and only one value of b.
4. k = { (x, y) / x³+y³=16xy } is not a function; It is not easy to show that this is the case. 

 In the following window we can see the graph of each of the four examples above. Use the vertical straight line joined to the control point (in red) to determine which are graphs of functions and which are not. 

1.- Use the window to prove these statements:

1. f = { (x, y) / y=2x } is a function.
2. g = { (a, b) / a²+b²=9 } is not a function.
3. h = { (a, b) / a²+b³=16 } is a function.
4. k = { (x, y) / x³+y³=16xy } is not a function. 

2.- Use the window to find three points belonging to the set in example 4 whose first coordinate is 8. The results you obtain will only be approximate. Use a calculator to check how good these results are.

When a function is defined by a property it is not necessarily defined by a single formula. It could be as complex as the following example, or perhaps even more so:

f₀ = { (x, y) / x²+y²=9 if y>=0; x²+y³=16 if y<0}

1.- Study the following examples which show how to calculate the domain and range:

• Let the function f = { (1, 2), (2, 4), (3, -1), (4, 2) }. Its domain is the set { 1, 2, 3, 4}. Its range is { 2, 4, -1 }. f(1)=2, f(2)=4, f(3)=-1 and f(4)=2.

• Let the function h = { (a, b) / a²+b³=16 } = { (x, y) / y=(16-x²)^1/3 }. By looking carefully at its graph in the first window we can deduce that its domain is all of R; its range is the set of numbers less than or equal to 16^1/3 (approximately 2.52); and, for example, h(-4)=0, h(0)=16^1/3.

2.- Find the domain and range of the function f₀ represented in the second window. It is worth looking back at the window to help you find the correct answer. What is the value of f₀(3), f₀(0), f₀(-1), f₀(-4), f₀(8) and f₀(3.5)?

We already know that a function is a set of pairs of numbers. If we reverse the numbers in the pairs we get a new function. Let's do so with the following function:

f = { (1, 2), (2, 4), (3, -1), (4, -2) }

We can see what happens and call this new set g: 

g = { (2, 1), (4, 2), (-1, 3), (-2, 4) }

We have obtained a new function.

However, this does not always work. Let f be the following set:

f = { (1, 2), (2, 4), (3, -1), (4, 2) }

Therefore g is:

g = { (2, 1), (4, 2), (-1, 3), (2, 4) }

which is not a function as g(2) is not determined by just one value; in other words g does not satisfy the condition of the function. There are two pairs of numbers, (2, 1) and (2, 4), whose first coordinate is the same and second coordinate is different.

What is the difference between these two examples? The reason is that in the second example f(1)=f(4)=2 and when the pairs are reversed g(2) is not determined by just one value; therefore g is not a function. In the first example the different values of "x" give different values of "y". Functions which behave in the same way as this first example are called injective or one-to-one.

When another function is obtained by reversing the pairs of a function we can say that this function has an inverse. In the light of what we have stated above, the only functions which have inverse functions are injective functions. 

From the definition we can see immediately that the domain of the inverse function f^-1 is the range of f and, reciprocally, the range of f^-1 is the domain of f. It is also easy to see that f^-1(a)=b is the equivalent of f(b)=a. By using "x" and "y" which we often use when referring to functions f^-1(x)=y is equivalent to f(y)=x. Another way of saying this is that: f(f^-1(x))=x (where x belongs to the range of f), or, f^-1(f(x))=x (where x belongs to the domain of f). By using the composition of functions, referring to the function defined by I(x)=x, I (Identity function) we can say that:

fof^-1 = I        and        f^-1of = I

except when the domain of the second member of these two equations is bigger than that of the first member if the domain of either f or f^-1 is not all of R.

Incidentally, if a function has an inverse, what will (f^-1)^-1 be equal to? i.e. the inverse of the inverse function?

We have often referred to the idea of inverse functions in earlier units at this level. However, we have not labelled them as such. Think back to how we defined the square root, cube root etc.

The following window represents the graphs of the two function f and g from activity 2. The functions are indicated in blue and the reverse pairs of numbers in pink. The control point moves along the functions indicating a pair of the function and the corresponding pair of the inverse function. The other straight line that you can see (in green) bisects the first and third quadrant (the straight line of the equation y=x). 

1.- Note that in order to determine whether a function has an inverse or not we must look at its pairs and see if it is an injective function. This is very straightforward when the function is given as a list of pairs. It is more complicated to do so if the function is defined by a property and we may not have enough mathematical knowledge to be able to decide (as was the case when we wanted to determine whether a certain set was a function or not). 

 If we have the graphical representation we can see that an injective function is defined graphically by the fact that no two points of the function are located on the same horizontal straight line. In other words, the graph of f is used to construct the graphical representation of the set of reverse pairs and this is used to see whether the set is a function or not.

2.- Analyse the following functions in the window:

1. The function f defined by y=2x-3, i.e., f = { (x, y) / y=2x-3 } = { (x, 2x-3) } has an inverse which is f^-1 = { (y, x) / y=2x-3 } = { (x, y) / x=2y-3 } = { (2x-3, x) }
2. The function g defined by y=x²-2x-2, i.e., g = { (x, y) / y=x²-2x-2 } = { (x, x²-2x-2) } does not have an inverse. For example, the pairs (0, -2) and (2, -2) belong to g and therefore, g is not injective.

3.- Does the function h = { (x, y) / y=x²-2x-2, if x > 1 } = { (x, x²-2x-2) / x > 1 } have an inverse? Use the window above to answer the question correctly. If the answer is yes, show the graphical representation of the inverse in the window by using the control button in the window. Does the function k = { (x, y) / y=x²-2x-2, if x < 1 } have an inverse?

4.- Note that the graphs of a function and the set of reverse pairs are symmetrical about the straight line y=x (in green).

The following window shows three functions.

2.- Use the controls in the window to construct the set of reverse pairs for each function.

3.- What happens with the third function? What can you say about the graph with respect to the straight line y=x? Write a summary of your observations and prove them if possible.

 We know that f^-1 = { (x, y) / x=f(y) } = { (f(y), y) / if y is the domain of f } = { (x, y) / y=f^-1(x), if x is the range of f }. However, what do we have to do to work out the equation for f^-1(x), i.e. what is the value of y as a function of the value of x when the pair (x, y) belongs to f^-1 ? The answer is simple: get y on its own in the equation x=f(y). Naturally, if x=f(y) is an equation, then if the function is given as a list of pairs calculations are unnecessary and if the function is given as a fairly complex expression we will need to examine it and decide if we can work out the equation of its inverse. We are going to use "simple" examples which will be based on examples we have already used.

Our problem is getting a value on its own. In algebraic equations we are able to get "letters" on their own on one side of the equation so we shouldn't have much difficulty in obtaining the equation f^-1(x) from the equation f(x).

The following window will allow us to check graphically that our calculations of the inverse function are correct when the function is given by a single equation. We start with the blue function; the pink function is the inverse which is expressed in the form x=f(y); the red function, shows the equation of the inverse function which we have obtained from getting y on its own in the equation x=f(y). All the equations should be used for each of the examples we want to check our answers to. Example 1 can be used as a model. If our calculations are correct then the groups of red and pink points should coincide. The red function has been written incorrectly in the window on purpose. Change the function and watch how it coincides with the pink one. The straight line equation y=x also appears in the window to emphasise how the graphs of a function and its inverse are symmetrical about this line.

1.- Analyse the following examples of inverse functions:

1. Let the function be f = { (x, y) / y=2x-3 }. Its inverse is: f^-1 = { (x, y) / x=2y-3 }. Get y on its own in the equation x=2y-3. Step by Step:   x+3=2y;   (x+3)/2=y. We now have f^-1(x)=(x+3)/2. Simple, as the equation of the function f was given by a polynomial of the first degree.
2. Let function g be g(x)=(-3x+5)/4. Its inverse g^-1 is g^-1(x)=(5-4x)/3. Follow the steps given in the example above to check that the solution is correct.

3. Let the function be f(x)=x²-2x-2. We have already seen that there is no inverse, but if we limit its domain to numbers greater than 1, then it does. In this case what is the equation of f^-1(x) ? We can find out by getting y on its own in the equation x=y²-2y-2 . The problem is getting y on its own; we still know enough to be able to do so as it is similar to getting y on its own in the equation y²-2y-2-x=0. In other words, solving a quadratic equation. We get two solutions using the traditional formula: y=1+(3+x)^1/2  ,  y=1-(3+x)^1/2. Which is the correct solution in this case? Find the correct answer by referring to the exercise (Function h = { (x, y) / y=x²-2x-2, if x > 1 } = { (x, x²-2x-2) / x > 1 } Does it have an inverse?) and by using the window above.
4. Let the function be f(x)=x³-2. Its inverse is: f^-1(x)=(x+2)^1/3. This is obtained by getting y on its own in the equation x=y³-2. Straightforward and easy to understand, I imagine.
5. Let the function be f(x)=x³+2x-2. Its inverse is: f^-1(x)=... Don't even try as we would need to solve a cubic equation to work it out, whose formula we haven't yet come across, although the formula does exist. Even so, draw f and f^-1in the window.

2.- Find the equation of f^-1(x) for the function f(x)=(2x+3)/(3x-6). It is the first function of section 4 in this unit. Remember that you can check the solution in the window above.