Every one-to-one function
$\,f\,$ has an
inverse;
this inverse is denoted by
$\,f^{-1}\,$
and read aloud as ‘$\,f\,$ inverse’.
A function and its inverse ‘undo’ each other:
one function does something, the other undoes it.
The purpose of this lesson is to make this idea precise.
The set of allowable inputs for a function $\,f\,$ is called its domain, and is denoted by $\,\text{dom}(f)\,$.
The set of all possible outputs from a function $\,f\,$ is called its
range,
and is denoted by $\,\text{ran}(f)\,$.
That is, let $\,f\,$ act on every possible inputevery element of its domain.
The set of resulting outputs is the range of $\,f\,$:
$$\cssId{s19}{\text{ran}(f)}
\cssId{s20}{= \{ f(x)\ |\ x\in\text{dom}(f) \}}
$$
In other words, the composite function $\,f^{-1}\circ f\,$ is the identity function on $\,\text{dom}(f)\,$. |
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In other words, the composite function $\,f\circ f^{-1}\,$ is the identity function on $\,\text{ran}(f)\,$. |
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The input/output roles for a function and its inverse are reversedthe inputs to one are the outputs from the other.
This fact has some nice consequences:
POINTS ON THE GRAPHS OF $\,f\,$ and $\,f^{-1}\,$ HAVE THEIR COORDINATES SWITCHED:
DOMAIN AND RANGE OF $\,f^{-1}\,$:
For your convenience, the properties of inverse functions discussed in this and earlier exercises are summarized below.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
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