Functions | Types, Examples & Summary | A Level Maths Revision Notes

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1 Summary

1.1 Domain and range

1.2 Example #1

1.3 One to One Function

1.4 Many to One Function

1.5 Composite Functions

1.6 Inverse Function

1.7 Example #2

1.8 Modulus Function

1.9 Graphical Transformations

1.9.1 Reference

Summary

A function is defined as a rule which assigns each element of set X, one and only one element of set Y. i.e set $X\quad =\quad \left\{ { x }_{ 1 },\quad x_{ 2 },\quad { x }_{ 3 } \right\}$ and set $Y\quad =\quad \left\{ { y }_{ 1 },\quad { y }_{ 2 },\quad { y }_{ 3 } \right\}$
Domain: elements of set X
Range: elements of set Y
One to One Function: for each element of set Y, there is a unique corresponding element in set X.
Many to One Function: for any element of set Y, there is more than one element in set X.
Inverse Function: $f^{ -1 }\left( y \right) \quad =\quad x$
Composite Function: combine two functions to get a new function $f\left[ g\left( x \right) \right]$
Modulus Function: $f\left( x \right) \quad =\quad \left| x \right|$

Let’s suppose we have two sets of numbers:

$X\quad =\quad \left\{ { 10,\quad 20,\quad 30,\quad 40 } \right\}$
$Y\quad =\quad \left\{ { 32,\quad 62,\quad 92,\quad 122 } \right\}$

To define a relationship between these two sets we write a function:

$f\left( x \right)\quad =\quad 3x\quad +\quad 2$ or can be written as $f\quad :\quad x\quad \rightarrow \quad 2x\quad +\quad 3$

Now we can see that for every value of set X, we have a value for set Y i.e

When $x\quad =\quad 10$ $f\left( x \right)\quad =\quad 3(10)\quad +\quad 2\quad =\quad 32$

When $x\quad =\quad 20$ $f\left( x \right) \quad =\quad 3(20)\quad +\quad 2\quad =\quad 62$ and so on…

If the following two conditions are satisfied than the relation in called a “Function”:

i) we can say that every element of X has an element in set Y.
ii) every element of X has only one unique element in set Y.

Therefore, a function is defined as a rule which assigns each element of set X, one and only one element of set Y.

Domain and range

Going back to the above example which consisted of two sets:

$X\quad =\quad \left\{ { 10,\quad 20,\quad 30,\quad 40 } \right\}$
$Y\quad =\quad \left\{ { 32,\quad 62,\quad 92,\quad 122 } \right\}$

Here elements of set X are known as “Domain” and elements of set Y are known as “Range”.

Hence we can write the domain and range as:

Domain of f = {10, 20, 30, 40}
Range of f = {32, 62, 92, 122}

Note: we cannot define a function if its domain is not mentioned, moreover range of a function may or may not consist of all elements of the set Y.

Example #1

Q. Let $f\quad :\quad x\quad \rightarrow \quad { (x\quad -\quad 4) }^{ 2 }\quad +\quad 8$ with domain x > 0.

Find the range of f.

Solution:

Since a function consists of a square, and we know that result of a square is always positive, therefore all values of f(x) will be greater than 8.

$f\quad :\quad x\quad \rightarrow \quad { (positive\quad value) }\quad +\quad 8$

Hence the range of f is > 8

One to One Function

If we have a function f such that for each element of set Y (the range), there is a unique corresponding element in set X (the domain), then f is called a one to one function i.e:

given $f\left( x \right)\quad =\quad 2x\quad +\quad 5$ from X to Y where X = {4, 7, 8} is one to one as shown below:

Many to One Function

If we have a function f such that for any element of set Y (the range), there is more than one element in set X (the domain), then f is called a many to one function, i.e:

$f\quad :\quad x\quad \rightarrow \quad 2{ x }^{ 2 }\quad +\quad 4x\quad -6$ is a many to one function as the result x = 1 & -3 both correspond to 0 of set Y.

Composite Functions

Suppose we have two functions:

$f\quad :\quad x\quad \rightarrow \quad { x }^{ 2 }$ and
$g\quad :\quad x\quad \rightarrow \quad { 2x\quad +\quad 3 }$

We can combine these two functions to get a new function $f\left[ g\left( x \right) \right]$ , and this new function is known as a composite function, i.e:

for the above two functions $f\left[ g\left( x \right) \right]$ means:

$=\quad f\left[ g\left( x \right) \right]$

$=\quad f\left( 2x\quad +\quad 3 \right)$

Since: $f\left( x \right) \quad =\quad { x }^{ 2 }$

$f\left[ g\left( x \right) \right] ={ (2x\quad +\quad 3) }^{ 2 }$

Inverse Function

Suppose we have a function $f\left( x \right)\quad =\quad y$ , inverse of this function is the one that uses reverse relation i.e

$f^{ -1 }\left( x \right) \quad =\quad y$

However, remember that an inverse of a function will only exist if a function is one to one. Furthermore, if we draw the graphs of $f\quad and\quad { f }^{ -1 }$ on the same axis, then the graph of ${ f }^{ -1 }$ will be the reflection of f in the line y = x as shown in Fig 1

Example #2

Q. $f\quad :\quad x\quad \rightarrow \quad \frac { 3x\quad +\quad 4 }{ 2 } ,\quad find\quad { f }^{ -1 }$

Solution:

We know:

$f\quad :\quad x\quad \rightarrow \quad \frac { 3x\quad +\quad 4 }{ 2 }$

$y\quad =\quad \frac { 3x\quad +\quad 4 }{ 2 }$

To find inverse make x the subject of equation:

$2y\quad =\quad 3x\quad +\quad 4$

$2y\quad -\quad 4\quad =\quad 3x$

$x\quad =\quad \frac { 2y\quad -\quad 4 }{ 3 }$

Hence,

${ f }^{ -1 }\quad \rightarrow \quad x\quad =\quad \frac { 2y\quad -\quad 4 }{ 3 }$ Ans

Modulus Function

A modulus is defined as an absolute value of x. It is basically a distance of x from a fixed point. It is written as $\left| x \right|$ . A modulus function is defined as:

Therefore the graph of $y\quad =\quad \left| x \right|$
is line y = x for x > 0 and y = -x for x < 0 as shown below in Fig 2:

Graphical Transformations

Suppose we have a function $y\quad =\quad f\left( x \right)$ , some rules to remember when transforming graphs are:

The graph of $y\quad =\quad f\left( x \right) \quad +\quad c$ where c is a constant, has the same shape of the function $f\left( x \right)$ but is moved c units higher.
The graph of $y\quad =\quad f\left( x \right) \quad -\quad c$ where c is a constant, has the same shape of the function $f\left( x \right)$ but is moved c units lower.
The graph of $y\quad =\quad f\left( x\quad +\quad k \right)$ , where k is a constant, has the same shape of the function but is moved k units to the left.
The graph of $y\quad =\quad f\left( x\quad -\quad k \right)$ , where k is a constant, has the same shape of the function but is moved k units to the right.
The graph of $y\quad =\quad f\left( -x \right)$ is simply a reflection of $f\left( x \right)$ in the y axis.
The graph of $y\quad =\quad -f\left( x \right)$ is a reflection of $f\left( x \right)$ in the x axis.
The graph of $y\quad =\quad a\quad f\left( x \right)$ is stretched by a scale factor a in the y axis.
The graph of $y\quad =\quad f\left( ax \right)$ is stretched by a scale factor $\frac { 1 }{ a }$ in the x axis.