A function is a mapping in which every input maps to exactly one output. This means one-one and many-one mappings are functions, but a mapping that sends a single input to more than one output is not.

What is the difference between a one-one mapping and a many-one mapping?

In a one-one mapping, every input has its own output, so no output value is ever repeated. In a many-one mapping, two or more different inputs give the same output. Both are functions, because in each case one input still gives only one output.

Why must a function be one-one for an inverse function to exist?

Because the inverse has to send each output back to a single input. If the function is many-one , two different inputs share an output, so the inverse would have to send that one value back to two places, and that is not a function.

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Functions (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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Define function.

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Define function.
A function is a mapping in which every input maps to exactly one output.
This means one-one and many-one mappings are functions, but a mapping that sends a single input to more than one output is not.
What is the difference between a one-one mapping and a many-one mapping?
In a one-one mapping, every input has its own output, so no output value is ever repeated.
In a many-one mapping, two or more different inputs give the same output.
Both are functions, because in each case one input still gives only one output.
A function is given by $f : x \mapsto 3 x - 4$ . What does this notation mean, and what is $f (5)$ ?
This is mapping notation, and it means the same as $f (x) = 3 x - 4$ .
To find $f (5)$ , replace every $x$ with 5:
$f (5) = 3 (5) - 4 = 11$
For the function $f (x) = 2 x + 1$ , the output is 15. How do you find the input?
Set the expression equal to the given output and solve the equation:
$2 x + 1 = 15 \Rightarrow x = 7$
Finding an output means substituting, but finding an input means solving.
Define the domain and the range of a function.
The domain is the set of values allowed as inputs, and the range is the set of all possible outputs.
On a graph, the domain values are the $x$ -coordinates and the range values are the $y$ -coordinates.
A function is written with no domain stated. What domain should you assume?
Assume the largest set of values the function can accept.
For example, $f (x) = \sqrt{x}$ has an assumed domain of $x \geq 0$ , because the square root of a negative number is not a real number.
Fill in the gaps for the function $f (x) = x^{2}$ :
With domain $- 10 \leq x \leq 10$ it is $______$ , but with domain $0 \leq x \leq 10$ it is $______$ .
The completed sentence is:
With domain $- 10 \leq x \leq 10$ it is many-one, but with domain $0 \leq x \leq 10$ it is one-one.
Restricting the domain removes the negative inputs, so no two inputs share an output any more.
The domain of $f (x) = 2 x + 1$ changes from $x \geq 0$ to $x \geq 10$ . What happens to the range?
The range changes from $f (x) \geq 1$ to $f (x) \geq 21$ , because the smallest input is now 10 and $2 (10) + 1 = 21$ .
Always work out the range from the domain you have been given, not from the expression alone.
Define the modulus of a number.
The modulus of a number is its size with any negative sign removed, so it is never negative.
It is written between vertical lines, so $|7| = 7$ and $|- 7| = 7$ .
It is also called the absolute value.
True or False?
If $f (x)$ is negative, then $|f (x)| = - f (x)$ .
True.
Multiplying a negative value by $- 1$ makes it positive, which is exactly what the modulus has to do.
For example, if $f (x) = - 7$ then $- f (x) = 7 = |f (x)|$ .
Define an inverse function.
An inverse function $f^{- 1}$ undoes whatever $f$ does, so applying $f$ and then $f^{- 1}$ returns the original input.
It carries out the inverse operations in the reverse order: if $f$ doubles and then adds 1, $f^{- 1}$ subtracts 1 and then halves.
Why must a function be one-one for an inverse function to exist?
Because the inverse has to send each output back to a single input.
If the function is many-one, two different inputs share an output, so the inverse would have to send that one value back to two places, and that is not a function.
Describe how to find $f^{- 1} (x)$ from $f (x) = 5 - 3 x$ .
Write it as $y = 5 - 3 x$ , swap the $x$ and the $y$ to get $x = 5 - 3 y$ , then rearrange to make $y$ the subject:
$f^{- 1} (x) = \frac{5 - x}{3}$
Finish in inverse-function notation, with no $y$ left in the answer.
$f (x) = 2^{x}$ . How can you solve $f^{- 1} (x) = 5$ without first finding $f^{- 1}$ ?
Apply $f$ to both sides and use the fact that ${ff}^{- 1} (x) = x$ :
$x = f (5) = 2^{5} = 32$
This works because a function and its inverse cancel each other out whenever they are applied together.
Fill in the gaps for a function $f$ and its inverse $f^{- 1}$ :
The domain of $f^{- 1}$ is the $______$ of $f$ , and the range of $f^{- 1}$ is the $______$ of $f$ .
The completed sentence is:
The domain of $f^{- 1}$ is the range of $f$ , and the range of $f^{- 1}$ is the domain of $f$ .
They swap over, because the inverse turns every output of $f$ back into the input it came from.
$f (x) = \sqrt{3 x - 2}$ for $x \geq \frac{2}{3}$ . Write down the domain and range of $f^{- 1} (x)$ .
The domain of $f^{- 1}$ is $x \geq 0$ , because that is the range of $f$ .
The range of $f^{- 1}$ is $f^{- 1} (x) \geq \frac{2}{3}$ , because that was the domain of $f$ .
How are the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ related?
Each is a reflection of the other in the line $y = x$ .
Every key feature reflects with it, so axis intercepts, turning points and asymptotes all swap their $x$ and $y$ roles.
The graph of $y = f (x)$ crosses the $x$ -axis at $(4, 0)$ and has the asymptote $x = - 2$ . What happens to each on the graph of $y = f^{- 1} (x)$ ?
The intercept becomes $(0, 4)$ and the asymptote becomes the line $y = - 2$ .
Reflecting in $y = x$ swaps every $x$ -coordinate with its $y$ -coordinate, which turns a vertical asymptote into a horizontal one.
True or False?
You need the expression for $f (x)$ before you can draw the graph of $y = f^{- 1} (x)$ .
False.
If you are given the graph of $y = f (x)$ , reflecting it in the line $y = x$ produces the graph of the inverse.
The reflection is a geometric operation, so it works whatever the expression for $f (x)$ happens to be.
Define a composite function.
A composite function applies one function to the output of another, so the output of the first becomes the input of the second.
$fg (x)$ means $f (g (x))$ , read as f of g of x.
In $fg (x)$ , which function is applied first?
$g$ is applied first, because it is the function closest to the variable.
Its output then becomes the input of $f$ , which is why $fg (x)$ means $f (g (x))$ .
$f (x) = x^{2}$ and $g (x) = 2 x$ . Find $fg (x)$ and $gf (x)$ , and say what they show.
$fg (x) = f (2 x) = (2 x)^{2} = 4 x^{2}$ , while $gf (x) = g (x^{2}) = 2 x^{2}$ .
They are different, which shows that the order matters: $fg$ is not usually the same as $gf$ .
True or False?
For some pairs of functions, $fg (x)$ and $gf (x)$ are exactly the same.
True.
The order usually matters, but not always.
For example, if $f (x) = 3 x - 2$ and $g (x) = 6 x - 5$ then $fg (x) = gf (x) = 18 x - 17$ .
What does $f^{2} (x)$ mean, and why is this notation not used with trigonometric functions?
$f^{2} (x)$ means $ff (x)$ , that is $f (f (x))$ .
It is not used with trigonometric functions because there the superscript means a power: $\sin^{2} (x)$ means $(\sin x)^{2}$ , not $\sin (\sin x)$ .
Fill in the gaps for the composite function $fg (x)$ :
The domain of $fg$ is at best the domain of $______$ , and the range of $fg$ is at best the range of $______$ .
The completed sentence is:
The domain of $fg$ is at best the domain of g, and the range of $fg$ is at best the range of f.
It is only at best because $fg$ needs the range of $g$ to fit inside the domain of $f$ , and where it does not, the domain of $fg$ has to be restricted.
$f (x) = \frac{1}{x}$ for $0 < x < 1$ , and $g (x) = x^{2}$ for $x > 1$ . Explain why $fg (x)$ does not exist.
The range of $g$ is $g (x) > 1$ , but the domain of $f$ is $0 < x < 1$ .
Nothing that $g$ produces is an acceptable input for $f$ , so $fg (x)$ cannot be formed.

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