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There is a symmetry between a function and its inverse. Specifically, if f is an invertible function with domain X and codomain Y, then its inverse f −1 has domain Y and image X, and the inverse of f −1 is the original function f. In symbols, for functions f:X → Y and f −1:Y → X, [13]
The graph of an involution (on the real numbers) is symmetric across the line y = x. This is due to the fact that the inverse of any general function will be its reflection over the line y = x. This can be seen by "swapping" x with y. If, in particular, the function is an involution, then its graph is its own reflection
The inverse of the curve C is then the locus of P as Q runs over C. The point O in this construction is called the center of inversion, the circle the circle of inversion, and k the radius of inversion. An inversion applied twice is the identity transformation, so the inverse of an inverse curve with respect to the same circle is the original ...
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse of f {\displaystyle f} is denoted as f − 1 {\displaystyle f^{-1}} , where f − 1 ( y ) = x {\displaystyle f^{-1}(y)=x} if and only if f ...
The graph of a function on its own does not determine the codomain. It is common [3] to use both terms function and graph of a function since even if considered the same object, they indicate viewing it from a different perspective. Graph of the function () = over the interval [−2,+3]. Also shown are the two real roots and the local minimum ...
For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth (1/5 or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4. The reciprocal function, the function f(x) that maps x to 1/x, is one of the simplest examples of a function which is its own inverse (an involution).
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In differential geometry, the inverse function theorem is used to show that the pre-image of a regular value under a smooth map is a manifold. [10] Indeed, let f : U → R r {\displaystyle f:U\to \mathbb {R} ^{r}} be such a smooth map from an open subset of R n {\displaystyle \mathbb {R} ^{n}} (since the result is local, there is no loss of ...