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.

Why must a function be one-to-one to have an inverse?

Because the inverse has to send each output back to a single input. If two inputs shared an output, the inverse would not know which of them to return to, and so would not itself be a function.

Two non-parallel straight lines meet once. Why can an equation involving a modulus have more solutions than that?

Because the modulus reflects part of a graph upwards, and the reflected piece can cross the other graph as well. That produces an intersection the unreflected lines never had.

What should you do before solving a modulus equation algebraically?

Sketch both graphs , including the reflected parts, and locate the intersections. The sketch tells you how many solutions there should be, and therefore which algebraic answers to keep.

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Functions (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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FrontLanguage of Functions
Define function.

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Cards in this collection (32)

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.
Define inverse function.
The inverse function $f^{- 1} (x)$ undoes what $f$ does, taking each output back to the input it came from.
It exists only if $f$ is one-to-one.
Why must a function be one-to-one to have an inverse?
Because the inverse has to send each output back to a single input.
If two inputs shared an output, the inverse would not know which of them to return to, and so would not itself be a function.
A function followed by its inverse returns the original input:
${ff}^{- 1} (x) = f^{- 1} f (x) =______$
${ff}^{- 1} (x) = f^{- 1} f (x) = x$
This is exactly what "undoing" means, and it holds for every value of $x$ in the domain.
How is the graph of $f^{- 1} (x)$ related to the graph of $f (x)$ ?
It is the reflection in the line $y = x$ .
Reflecting in that line swaps the two coordinates of every point, which is precisely what taking the inverse does.
True or False?
The domain of $f^{- 1}$ is the range of $f$ .
True.
The two swap over, because the outputs of $f$ are exactly the inputs of $f^{- 1}$ .
The domain of $f$ is likewise the range of $f^{- 1}$ .
How do you find an inverse function algebraically?
Write $y = f (x)$ , then rearrange to make $x$ the subject.
Rewrite the result in function notation, remembering to give its domain as well.
How can an inverse function be used to solve $f (x) = 5$ ?
The solution is $x = f^{- 1} (5)$ .
Applying $f^{- 1}$ to both sides undoes $f$ on the left, leaving $x$ on its own.
Define the modulus of a number.
The modulus of a number is its size with the sign removed, so $|5| = 5$ and $|- 5| = 5$ .
Formally, $|x| = x$ when $x \geq 0$ and $|x| = - x$ when $x < 0$ .
It is also called the absolute value.
For $y = a |x + p| + q$ , what shape is the graph and where is its vertex?
A " $\lor$ " shape if $a > 0$ and a " $\land$ " shape if $a < 0$ , with the vertex at $(- p, q)$ .
The form behaves just like the completed square $a {(x + p)}^{2} + q$ , whose turning point sits in the same place.
How do you sketch $y = |f (x)|$ from the graph of $y = f (x)$ ?
Sketch $y = f (x)$ first, then reflect everything below the $x$ -axis up into it.
Anything already on or above the axis stays exactly where it is.
How do you sketch $y = f (|x|)$ from the graph of $y = f (x)$ ?
Sketch $y = f (x)$ only for $x \geq 0$ , then reflect that part in the $y$ -axis.
Whatever the original graph did for negative $x$ is discarded.
True or False?
The graph of $y = f (|x|)$ can go below the $x$ -axis.
True.
Only $y = |f (x)|$ is forced to stay on or above the axis, because there it is the output that is made positive.
$y = f (|x|)$ makes the input positive instead, so its outputs can still be negative.
Which of $y = |f (x)|$ and $y = f (|x|)$ is always symmetrical, and about what?
$y = f (|x|)$ is always symmetrical about the $y$ -axis, because $|x|$ and $|- x|$ are equal.
$y = |f (x)|$ need have no line of symmetry at all.
The graph of $y = a |x + p| + q$ can have $______$ , $______$ or $______$ roots.
The graph of $y = a |x + p| + q$ can have 0, 1 or 2 roots.
Which it is depends on where the vertex sits and whether the graph opens upwards or downwards.
Two non-parallel straight lines meet once. Why can an equation involving a modulus have more solutions than that?
Because the modulus reflects part of a graph upwards, and the reflected piece can cross the other graph as well.
That produces an intersection the unreflected lines never had.
To solve $|f (x)| = |g (x)|$ , solve both $f (x) = g (x)$ and $f (x) =______$ .
To solve $|f (x)| = |g (x)|$ , solve both $f (x) = g (x)$ and $f (x) = - g (x)$ .
The two cases arise because each side can take either sign once the modulus is removed.
What should you do before solving a modulus equation algebraically?
Sketch both graphs, including the reflected parts, and locate the intersections.
The sketch tells you how many solutions there should be, and therefore which algebraic answers to keep.
True or False?
Every solution of $x - 4 = 2 x - 5$ is also a solution of $|x - 4| = 2 x - 5$ .
False.
$x = 1$ satisfies the first, but $|1 - 4| = 3$ while $2 (1) - 5 = - 3$ , so it fails the second.
How do you solve $|f (x)| = g (x)$ , with a modulus on one side only?
Solve both $f (x) = g (x)$ and $- f (x) = g (x)$ .
Then test each answer in the original equation, because a modulus can never be negative, so any root making $g (x)$ negative has to be thrown out.

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