Inverse functions. Thorough study of concept.

Mathematical Software. Mathematical Research. Mathematical Education. Tvalx Products.



In the article Exploring Inverse Functions we made first approach to the topic avoiding formal definition of inverse function. Let's study thoroughly the concept of inverse function.

From set-theoretical point of view a function is a set of ordered pairs. For example, y=2x is {(x, 2*x) | x ϵ R} . Take any one-to-one function f = {(x, y) | x ϵ D, y=f(x)}, where f denotes the function as a set of ordered pairs and as a rule of computing y for given x. The domain of f is D and the range is R. Consider set of ordered pairs g = {(y, x) | x ϵ D, y=f(x)}. So, we switched x and y. Is g a function? Since f is one-to-one, yes, g is a function. For each given y we have exactly one corresponding x. The letters x and y are just symbols used inside curled brackets. If we write  g = {(a, b) | x ϵ D, a=f(b)}, the logic does not change. So we can write g = {(x, y) | y ϵ D, x=f(y)} to follow convention that argument is denoted by x and value is denoted by y. Thus we can formulate a definition:

Definition 1.

Let  f = {(x, y) | x ϵ D, y=f(x)} be a one-to-one function. Then g = {(x, y) | y ϵ D, x=f(y)}is called inverse function for f.


By convention an inverse function of f is denoted by f -1 . So g = f -1 .

It's possible to give a strict set-theoretical proof that there is exactly one inverse function for given f. Let's accept it as obvious.

Note that if g is the inverse function for f then f is the inverse function for g. Indeed, f = {(x, y) | x ϵ D, y=f(x)} = {(x, y) | y ϵ R, x=g(y)} . Thus the logical roles of f and g are symmetrical. Neither f nor g have any preference over each other. Thus we can say that for any one-to-one function there is exactly one pair of mutually inverse functions. Indeed, switching x and y results in switching between two functions. Let's call it a pair of mutually inverse functions.

Among standard functions we have following pairs of mutually inverse functions:  (e^x, ln(x)), (sinh(x), asinh(x)) . We can also construct more trivial pairs of mutually inverse functions: (2x, x/2), (x^3, x^(1/3)) . During such construction we "solve" original function for y. For example, y = 1 + 2x is solved for y : x = (y -1) / 2 . Thus we get a pair of mutually inverse functions (1 + 2x, (x - 1) /2) . We should use such "solution" carefully, since it works only for one-to-one functions (one-to-one functions are called bijections in Set Theory).

We used to think of arcsine as inverse function of sine. Is it not? In sense of Definition 1 the answer is No. Arcsine is a bijection and has an inverse function, that is a restriction of sine to interval [-π/2, π/2] . Thus arcsine and restriction of sine to interval [-π/2, π/2]  form a pair of inverse functions. We could call sine and arcsine a conventional pair of inverse functions. Another example of conventional pair of inverse functions is a pair of x squared and square root of x. But we should remember that facts, which are true for pairs of inverse functions, are not necessarily true for conventional pairs of inverse functions.


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