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Revision:OCR Core 3 - Functions

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 3 - Functions

1 2. Functions
2 Also See
3 Comments

2. Functions

The domain of a function

Some functions cannot be represented in the sense that it is not possible to show every value for which the function is defined (the graph of $y = x^{5}$ is defined for all real values of x, and hence it cannot be shown in full). Some other functions are not like this; consider a square root; the square root of a number is not defined (as a real number) for negative values, hence there are some values for which these functions have no defined value.

Sometimes values for which a function is undefined (mathematically) are not the only restrictions. Consider a function which calculates the volume of a cube, of side length "x" ( $f(x) = x^{3}$ ). It is not valid for this function to be defined for any negative value of "x", as a length cannot be negative.

The domain of a function is simply the set of numbers for which that function is defined (and applicable). Hence, the function of "x" $f(x) = \frac{1}{x}$ is defined for all real values of "x", except 0; hence the domain is $x \not = 0$ .

The range of a function

As the values which can be input into a function can be restricted, it is interesting to consider how the value that the function can take (or output) can be limited.

Functions such as $f(x) = x^{2}$ are limited in which values they can take; for instance, there is no value of "x" in the domain of this function which would produce a value of less than zero when the function was applied.

Example

1. Calculate the domain and range of the function of "x": $f(x) = x^{2} + 3x + 4$ .

The range for this function is all real "x" ( $x \in \mathbb{R}$ ), as there are no real value of "x" for which any aspect of the function is undefined.

This question can be approached using differential calculus. Consider that this is a quadratic function whose highest power term has a positive coefficient, meaning that there would be a minimum value if it were to be plotted. The reason behind considering the function as a graph is simply that this allows the representation of a large number of values of "x" in an intuitive manner.

$f^{'}(x) = 2x + 3$

Hence the minimum is at: $x = - \frac{3}{2}$ .

Calculate the corresponding value of the function: $f(x) = \frac{9}{4} - \frac{9}{2} + \frac{8}{2} = \frac{7}{4}$ .

Hence the range is: $f(x) \ge \frac{7}{4}$ .

Function notation

Functions are like a series of operations, and as such they can be represented in a flow diagram, with arrows linking the different operations of the function. These diagrams are useful for finding the inverse function (later in this section of the notes), but are generally rather large and cumbersome.

Instead, the following notation is used:

$f : x \mapsto f(x)$

This shows that the function "f" maps all "x" in the domain onto a value of "f(x)" in the range. For example: $f : x \mapsto x^{2} + 2x + 4$ is the function which maps all real "x" onto a value "f(x)", such that $f(x) \ge 3$ .

Forming composite functions

The idea of the flow diagram can be brought into the functions. For instance, if $f(x) = (x + 2)^{2} + 4$ , finding f(8) would be done as follows:

$8 \rightarrow [add \ 2] \rightarrow 10 \rightarrow [square] \rightarrow 100 \rightarrow [add \ 4] \rightarrow 104$ .

Within this function there are several other functions, "add 2", "square", "add 4"; this means that the function "f" is called a composite of the aforementioned functions, with the components being the aforementioned functions.

This can be shown more generally:

$x \rightarrow [f_{1}] \rightarrow f_{1}(x) \rightarrow [f_{2}] \rightarrow f_{2}(f(x)) \rightarrow ... \rightarrow [f_{n}] \rightarrow f_{n}(f_{n - 1}( ... f_{2}(f_{1}(x)))...))$ .

If the brackets (on the end expression) are removed, the remaining expression can be used, however this shows that the most inner function ( $f_{1}$ ) is applied first (not the most outer). The order is important, as $f_{2}f_{1}(x) \not = f_{1}f_{2}(x)$ for a lot of functions.

Domain and range revisited

In Core 3 the functions used are defined only by real numbers, however it has already been demonstrated that the numbers for which a function is defined will be limited (for some functions). The natural domain refers to the numbers for which the function will give an answer (so for the example of restrictions based upon the subject of the question the natural domain is not being used).

In some cases, a composite function will not give defined values for numbers which are in the domain and range of the components:

$f : x \mapsto \cos ^{-1} ( x + 2 )$ .

Both of the components in the above function are defined for x = 1 , however f(1) is not defined.

Consider a general composite function (with two components):

$gf : x \mapsto g(f(x))$ .

This implies that the output of the function "f" is the input of the function "g", and therefore the range of "f" must be a subset of the domaim of "g" in order for all values in the domain of "f" to produce defined values by "gf".

In the previous general example, the problem lies in that the domain of $\cos ^{-1}$ is $-1 \le \cos ^{-1} (x) \le 1$ , and the range of the function x + 2 is $(x + 2) \in \mathbb{R}$ . Hence there are values of the first function for which the second is not defined.

Reversing functions

Most people are aware of inverse operations, however this idea can be extended to a series of operations, and therefore functions. The inverse function (with respect to some function) is merely a function such that all values in the domain are mapped onto the values which produced them by the original function. Consider the easy example:

$f : x \mapsto x + 10$

This function has a single operation, so the inverse is easily defined as:

$f^{-1} : x \mapsto x - 10$

f(4) = 14

$f^{-1} (14) = 4$ .

So, in general, the inverse function is a function $f^{-1}$ such that for a function , with domain , $\forall x \in D, \ f^{-1}f(x) = x$ .

Note that some functions will invert themselves, such as "divide by -1".

One-one functions

For some functions there are several values which are mapped onto the same value by the function; the trigonometric functions are good examples of these. These functions are not one-one functions with their natural domain as there is more than one value that will map onto a given value. As such, inverse functions are not defined for those functions that are not one-one (as there would be ambiguity as to which value would be meant; functions could have their inverses applied and produce different answers [against the idea of an inverse]).

The identity functions are $ff^{-1}$ , and $f^{-1}f$ due to the fact that these both produce the output which is the same as the input; however there may be implications on certain values being defined (due to the order).

Finding inverse functions

In the previous example, finding the inverse was very easy, as it is simple to see that the single operation is inverted by using the inverse operation. In a more complex situation a flow diagram might be of use, for example:

$f : x \mapsto 2x + 3$ .

First this can be represented as a flow diagram:

$x \rightarrow [double] \rightarrow 2x \rightarrow [add \ 3] \rightarrow 2x + 3$ .

It will probably be obvious that working backwards will produce the inverse function:

$2x + 3 \rightarrow [subtract \ 3] \rightarrow 2x \rightarrow [halve] \rightarrow x$ .

It is simple to explain why this works in a more general sense.

For functions which can be defined in this manner, there is simply a series of operations being carried out; consider these as functions, hence:

$f(x) = g_{n}g_{n - 1}...g_{2}g_{1}(x)$ .

This implies that the first function applied was $g_{1}$ , however, to remove these functions (invert the function "f"), $g_{n}$ must be inverted first, followed by $g_{n - 1}$ , and so on. The last function to be applied must be the first to be removed.

Another method is useful in cases where it is not possible to use this flow diagram method.

f(x) = y

Therefore:

$f ^{-1}(y) = x$ .

Therefore it is sufficient to find some function of "y" which maps onto "x".

Example

1. Find $f^{-1}$ if $f : x \mapsto \frac{x + 3}{2x + 3}$ (over a suitable domain).

$\frac{x + 3}{2x + 3} = y \\ \therefore x + 3 = 2xy + 3y \\ \therefore 3 - 3y = 2xy - x \\ \therefore 3 - 3y = x(2y - 1) \\ \therefore x = \frac{3 - 3y}{2y - 1}$ .

Hence:

$f ^{-1} : x \mapsto \frac{3 - 3x}{3x - 1}$

(Note that the letter used does not matter, and often it is more understandable to use the letter used in the original question to defined the inverse function [it is also more economical in the context of use of letters]).

In some cases where a quadratic is solved (in order to obtain the inverse function) the plus or minus symbol must be dealt with. Whether the plus or the minus is chosen will depend on the original function.