Inverse function

In mathematics, an inverse function is in simple terms a function which "does the reverse" of a given function. More formally, if f is a function with domain X, then f⁻¹ is its inverse function if and only if for every $x\in X$ we have:

f^{-1}(f(x))=f(f^{-1}(x))=x.\,

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

f\colon x\to 3x+2\,

f^{-1}\colon x\to (x-2)/3\,

If a function f has an inverse then f is said to be invertible. Also, if an inverse exists, it is unique.

The superscript "−1" is not an exponent. Similarly, as long as we are not in trigonometry or calculus, f ²(x) means "do f twice", that is f(f(x)), not the square of f(x). For example, if : f : x → 3x + 2, then f ² : x = 3 ((3x + 2)) + 2, or 9x + 8. However, in trigonometry, for historical reasons, sin²(x) usually does mean the square of sin(x). As such, the prefix arc is sometimes used to denote inverse trigonometric functions, e.g. arcsin x for the inverse of sin(x). In calculus, f⁽ⁿ⁾(x) is the nth derivative of f.

Simplifying rule

Generally, if f(x) is any function, and g is its inverse, then g(f(x)) = x and f(g(x)) = x. In other words, an inverse function undoes what the original function does. In the above example, we can prove f⁻¹ is the inverse by substituting (x − 2) / 3 into f, so

3(x − 2) / 3 + 2 = x.

Similarly this can be shown for substituting f into f⁻¹.

Indeed, an equivalent definition of an inverse function g of f, is to require that g o f be the identity function on the domain of f, and f o g be the identity function on the codomain of f, where "o" represents function composition.

Existence

For a function f to have a valid inverse, it must be a bijection, that is:

(f is onto) each element in the codomain must be "hit" by f: otherwise there would be no way of defining the inverse of f for some elements.
(f is one-to-one) each element in the codomain must be "hit" by f only once: otherwise the inverse function 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

y=f^{-1}(x)\,

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.
$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$ , provided all indicated compositions and inverses exist.
The inverse function and the inverse image of a set coincide in the following sense. Suppose $f^{-1}(A)$ is the inverse image of a set $A\subset Y$ under a function $f:X\to Y.$ If $f$ is a bijection, 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.
$(f^{-1})'={\frac {1}{f'\circ f^{-1}}}$ , for any differentiable function f of a real argument and with real values, when the indicated compositions and inverses exist.

Series form

Inverse function, if exist, can be calculated as a sum of the following series:

$f^{-1}=\sum _{k=0}^{\infty }A_{k}{\frac {(t-f)^{k}}{k!}},{\mbox{ where }}\left\{{\begin{matrix}A_{0}=t&\\A_{n+1}={\frac {A_{n}}{f'}}&\end{matrix}}\right.$

Left inverses, right inverses, and partial functions

A function f has at least one "left inverse" if and only if it is an injection. A left inverse is a function g such that
$g(f(x))=x$ .
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.

The same way, f has at least one "right inverse" ( $f(g(x))=x$ ) if and only if it is a surjection. 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 $\mathbb {R}$ to $\mathbb {R} ^{+}$ . 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.

Simplifying rule

Existence

Properties

Series form

Left inverses, right inverses, and partial functions

See also