Which types of mapping are functions?

One-to-one and many-to-one mappings. One-to-many and many-to-many are not, because a single input would have more than one output.

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Functions (Cambridge (CIE) A Level Maths: Pure 1): Flashcards

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FrontLanguage of Functions
Define function.

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Cards in this collection (13)

Define function.
A function is a mapping in which every 'input' maps to exactly one 'output'.
Functions are written as $f (x)$ , or in the alternative form $f : x \mapsto x^{2} - 3 x + 2$ .
Which types of mapping are functions?
One-to-one and many-to-one mappings.
One-to-many and many-to-many are not, because a single input would have more than one output.
The $______$ of a function is the set of allowed inputs, and the $______$ is the set of all possible outputs.
The domain of a function is the set of allowed inputs, and the range is the set of all possible outputs.
The range depends on the domain: change which inputs are allowed and the outputs change with them.
Why is squaring a many-to-one mapping?
Because different inputs can give the same output: both $2$ and $- 2$ map to $4$ .
It is still a function, since each individual input has only one output.
True or False?
$f (x) = x^{2}$ on its own is a fully defined function.
False.
A function is not fully defined until its domain has been stated.
Without it you cannot say what the range is, or whether the function is one-to-one.
How can restricting the domain change a function?
It can turn a many-to-one function into a one-to-one one.
$f (x) = x^{2}$ is many-to-one over all real $x$ , but restricting it to $x \geq 0$ makes it one-to-one.
How do you work out the range of a function?
Find the outputs that the stated domain actually produces, sketching the graph where that helps.
For $f (x) = x^{2}$ with $x \in ℝ$ the range is $f (x) \geq 0$ , because a square is never negative.
Define composite function.
A composite function is one function applied after another, with the output of the first becoming the input of the second.
It can be written $fg (x)$ , $f (g (x))$ or $(f \circ g) (x)$ , all meaning " $f$ of $g (x)$ ".
In $fg (x)$ , which function is applied first?
$g$ , the one closest to the variable.
You work outwards: apply $g$ to $x$ first, then apply $f$ to whatever that produces.
True or False?
$fg (x)$ and $gf (x)$ always give the same result.
False.
The order in which the functions are applied changes the answer, so the two are usually different.
They can happen to agree for particular pairs of functions, but that cannot be assumed.
Applying a function to its own output, $ff (x)$ , is written as $______$ .
Applying a function to its own output, $ff (x)$ , is written as $f^{2} (x)$ .
This means apply $f$ twice, not square the answer.
Why might $fg (x)$ exist when $gf (x)$ does not?
Because the output of the first function has to be an allowed input for the second.
The range of $g$ must lie inside the domain of $f$ , and that can hold one way round without holding the other.
Given $f (x) = 2 x$ and $g (x) = x + 3$ , what is $fg (x)$ ?
$fg (x) = 2 (x + 3) = 2 x + 6$ .
Taking them the other way round gives $gf (x) = 2 x + 3$ , which is a different function.

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