Lesson INTRODUCTION TO FUNCTIONS

Source code of 'INTRODUCTION TO FUNCTIONS'

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If there are two variables such that if one (called {{{red(Independent)}}}{{{red(Variable)}}}) changes, value the other (called {{{red(Dependent)}}}{{{red(Variable)}}}) also changes its value then the relationship between the two variables is called {{{red(Function)}}}. This means that these two variables are linked in such a way that it is imposible to change the value of one keeping the other fixed at a certain value.

Let 'x' and 'y' be such two variables. Now, y = f(x) means 'y' expressed as function of 'x' that is this relation gives a value (or may also give more than one value) for a single value of 'x'. Accordingly the 'y' is called {{{red(Single)}}}{{{red(Valued)}}}{{{red(Function)}}} (if one value of 'x' corresponds to only one value of 'y'; {{{y=2x}}}, for x = 4, y = 8, thus only one value of 'y' for one value of 'x') or {{{red(Multiple)}}}{{{red(Valued)}}}{{{red(Function)}}} (if one value of 'x' corresponds to more than one value of 'y'; {{{y = sqrt(x)}}}, for x = 4, y = 2 or -2, thus two values for one value of 'x') of 'x'.

A function on a plane means a function involving two variables because two values (abscissae and ordinate in Cartesian reference frame; radius vactor and its inclination angle in Polar reference frame; etc.) are sufficient to locate a point uniquely in a plane. A plane function represents either a curve or a straight line. The {{{red(Domain)}}} of the function is the range of values within which the abscissae lies and the {{{red(Range)}}} of the function is the range of values within which the ordinate lies.

ILLUSTRATION
# Find the domain and range of the function {{{f(x) = 1/sqrt(4-x^2)}}}.

Solution:

DOMAIN OF DEFINITION
For f(x) to be defined, the denominator must not be zero.
So 'x' cannot be equal to 2 or -2.
Also, the expression under the square root have to be positive.
So {{{4-x^2 > 0}}}
or {{{(2+x)(2-x) > 0}}}
Therefore either both (2 + x) and (2 - x) greater than zero or both are less than zero.
In first case, 
{{{(2 + x) > 0}}} or {{{x > -2}}} and {{{(2 - x) > 0}}} or {{{x < 2}}}; so {{{-2 < x <2}}}.
In second case, 
{{{(2 + x) < 0}}} or {{{x < -2}}} and {{{(2 - x) < 0}}} or {{{x > 2}}}; 'x' cannot be less than -2 but greater than 2 at same time. So the second case is inadmissible.
Hence the domain of definition of f(x) is (-2,2).

N.B. Here (-2,2) means {{{-2 < x < 2}}}. Had it been {{{-2 <= x <= 2}}} then it would have been denoted by [-2,2], had it been {{{- 2 <= x < 2}}} it would have been denoted by [-2,2) and (-2,2] in case of {{{-2 < x <= 2}}}.

RANGE OF THE FUNCTION
To find a range of a function y = f(x), we have to first find the inverse function of f(x) i.e. we have to express 'x' in terms of 'y' and then find the domain of definition of this inverse function. This is the range of the function f(x).
Here {{{y = 1/sqrt(4-x^2)}}}
or {{{y^2 = 1/(4-x^2)}}}
or {{{(4-x^2) = 1/y^2}}}
or {{{x^2 = 4 - 1/y^2}}}
or {{{x = (sqrt(4y^2 - 1))/y}}}
Now, find the domain of this function in 'y'.
Do this part yourself and check with the answer.
Answer: The range is ({{{-infinity}}},{{{-1/2}}}] and [{{{1/2}}},{{{infinity}}}).

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