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DEFINITION AND NOTATION

Formally, if ''f'' is a function with Domain ''X'' and Codomain ''Y'' (i.e., f \colon X o Y), then ''f'' -1 is its inverse function If And Only If both of the following conditions are met:

# for every x \in X we have f^{-1}(f(x)) = x \, (''f'' can be undone by ''f'' -1),
# for every y \in Y we have f(f^{-1}(y)) = y \, (''f'' -1 can be undone by ''f'' ).

This implies that f^{-1} \colon Y o X.

For example, if the function ''x'' → 3''x'' + 2 is given, then its inverse function is ''x'' → (''x'' −2) / 3. This is usually written as

: f\colon x o 3x + 2,
: f^{-1}\colon x o(x-2)/3.

An inverse function undoes what the original function does, and vice versa. In the above example, we can prove ''f'' -1 is the inverse of ''f '' by substituting (''x'' − 2) / 3 into ''f'':
: 3 − 2) / 3 + 2 = ''x''.

Similarly it can be shown by substituting ''f'' into ''f'' -1 that

: + 2) - 2 / 3 = ''x''.

If a function ''f'' has an inverse, then ''f'' is said to be invertible. If an inverse exists, it is unique. Most functions encountered in elementary calculus do not have an inverse.Smith, William K. ''Inverse Functions'', MacMillan, 1966 (p. 60).

The superscript "−1" is not an exponent. Similarly, except when dealing with s, e.g., arcsin ''x'' for the inverse of sin(''x''). In calculus, ''f'' (''n'')(''x'') is the ''n''th Derivative of ''f''.


EQUIVALENT DEFINITIONS


The above definition can be written in different equivalent forms. Three examples are given below.


Using more concise terminology

A function f\colon X o Y is invertible If And Only If an inverse function ''g'' exists such that:
# g(f(x)) = x \, for all ''x'' in ''X'',
# f(g(x)) = y \, for all ''y'' in ''Y''.


Using function composition

A function f\colon X o Y is invertible If And Only If an inverse function ''g'' exists such that:
# (g \circ f)(x) = x \, for all ''x'' in ''X'' (''g'' o ''f'' is the Identity Function on ''X''),
# (f \circ g)(y) = y \, for all ''y'' in ''Y'' (''f'' o ''g'' is the identity function on ''Y''),
where "o" represents Function Composition .


Using symbolic logic

A function f\colon X o Y is invertible If And Only If an inverse function ''g'' exists such that:
# orall{x}{\in} X\,\, (g \circ f)(x) = x ,
# orall{y}{\in} Y\,\, (f \circ g)(y) = y ,
where orall is the Universal Quantifier , and "o" represents Function Composition .


EXISTENCE

A function ''f'' has an inverse if and only if it is a Bijection . Consider:
  • ''f'' must be Onto (surjective); each element in the Codomain must be "hit" by ''f'': otherwise there would be no way of defining the inverse function ''f'' −1 for some elements.

  • ''f'' must be One-to-one (injective); each element in the codomain must be "hit" by ''f'' only once: otherwise the inverse function ''f'' −1 would have to send that element back to more than one value.


If ''f'' is a real-valued function, then for ''f'' to have a valid inverse, it must pass the Horizontal Line Test , that is a horizontal line ''y'' = ''k'' placed on the graph of ''f'' must pass through ''f'' exactly once for all real ''k''.

It is possible to work around this condition, by redefining ''f'''s codomain to be precisely its Range , and by admitting a multi-valued function as an inverse.

If one represents the function ''f'' graphically in an ''x''-''y'' Coordinate System , then the graph of ''f'' −1 is the reflection of the graph of ''f'' across the line ''y'' = ''x''.

Algebraically, one computes the inverse function of ''f'' by solving the equation
: y=f(x) \,
for ''x'', and then exchanging ''y'' and ''x'' to get
:x=f^{-1}(y) \,
This is not always easy; if the function ''f''(''x'') is Analytic , the Lagrange Inversion Theorem may be used.

The symbol f −1 is also used for the (set valued) function associating to an element or a subset of the codomain, the Inverse Image of this subset (or element, seen as a Singleton ).


PROPERTIES

  • When an inverse function exists, it is unique.

  • If ''f'' is invertible, its range ''R'' coincides with its codomain ''Y'' (an immediate consequence of the fact that all invertible functions are Surjective ).

  • If ''f'' is invertible, its codomain ''Y'' has the same Cardinality as its domain ''X'', or exactly the same size as ''X'', if ''X'' is a Finite Set (an immediate consequence of the fact that all invertible functions are Bijective ).

  • (f\circ g)^{-1}=g^{-1}\circ f^{-1}, provided all indicated compositions and inverses exist.

  • The inverse function and the , then f^{-1}(y)=f^{-1}(\{y\}).

  • A Linear Mapping between Vector Spaces is invertible if and only if the Determinant of the mapping is nonzero.

  • For a function between Euclidean Spaces , the Inverse Function Theorem gives a sufficient condition for the function to have a locally defined inverse.

  • (f^{-1})'= rac{1}{f'(f^{-1})}, for any differentiable function ''f'' of a real argument and with real values, when the indicated compositions and inverses exist.

  • If f(x)=y, then f^{-1}(y)=x

  • Akin to the fact that for a graphed function f, it may not produce different results for the same input, the function f^{-1} may not produce the same value for different inputs.



LEFT INVERSES, RIGHT INVERSES, AND PARTIAL FUNCTIONS

A function ''f'' with non-empty domain has at least one "left inverse" if and only if it is an Injection . A left inverse is a function ''g'' such that, for every x \in X

:g(f(x)) = x \, (''f'' can be undone by ''g'')

If ''f'' is not a surjection, we obtain ''g'' by setting g(f(x)) = x \, for each element in the range of ''x'', and g(y) = z \,, where ''z'' is any element whatever, for any ''y'' in the codomain of ''f'' but not in its range.

Similarly, ''f'' has at least one "right inverse" if and only if it is a Surjection . A right inverse is a function ''g'' such that, for every y \in Y

:f(g(y)) = y \, (''g'' can be undone by ''f'')

Here, for each ''x'', ''g'' assigns one of the elements in the domain of ''f'' which "produce" ''x''. For example, we know that f(x) = x^2 \, is a surjection from \R to \R^+. Then, g(x) = -\sqrt{x} is a famous right inverse to x^2, because (-\sqrt{x})^2 = x for all x \in \R^+. But it is not a left inverse: -\sqrt{x^2} = -x for x \in \R^+.

If ''f'' is a bijection, then the (unique) right inverse equals the left inverse, and we have come again to the ordinary inverse described above.

Using this definition, we can view any Partial Function as a left inverse of an injection. Because the Range of a left inverse is not restricted, we can adjoin to the domain of this injection an element "undefined", which we then assign to every element of the codomain which is not in the range.


SEE ALSO