Introduction Of Functions

DEFINITION

Function can be easily defined with the help of the concept of mapping. Let X and Y be any two non-empty sets. "A function from X to Y is a rule or correspondence that assigns to each element of set X, one and only one element of set Y". Let the correspondence be 'f' then mathematically we write f: X$\to$ Y where y = f(x), x $\in$X and y $\in$Y. We say that 'y' is the image of 'x' under 'f ' (or x is the pre image of y).

$\bullet$ A mapping f: X$\to$Y is said to be a function if each element in the set X has it's image in set Y. It is possible that a few elements in the set Y are present which are not the images of any element in set X.

$\bullet$ Every element in set X should have one and only one image. That means it is impossible to have more than one image for a specific element in set X. Functions can't be multi-valued (A mapping that is multi-valued is called a relation from X to Y)

Illustration 1.

Let X = {1, –1, 2, 3, –3}, Y = {1, 4, 9, 10,11}. The Rule $f$ given by y = x² is a function from X to Y. Domain $f$ = X = {1, –1, 2, 3, –3} Range $f$ = {1, 4, 9}

Illustration 2.

Let X = {1, –1, 2, 3}, Y = {2, –2, 4, 6}. The rule $f$ given by y = 2x is a function from X to Y. Domain = $f$ = X = {1, –1, 2, 3}, Range = {2, –2, 4, 6}

Illustration 3.

Let A = {1, 4, 9}, B = {-3, - 2, -1,1, 2, 3, 4}. The rule $f$ given by y² = x is not a function.

Graphical method: to check whether a relation between x and y is a function or not, draw a line parallel to y-axis and, if it interest the graph at one and only one point then the given relation represent a function.

VALUE OF A FUNCTION

The value of a function y = $f$(x) at x = a is denoted by $f$(a). It is obtained by putting x = a

in $f$(x)

Illustration 4 If $f\text{(x)=}log\left(\frac{\text{1-x}}{\text{1+x}}\right)$, x$\in$(–1, 1), then find the value of function for x=0,1/2.

Solution. $f\text{(0) = }[f(x){]}_{x=0}\text{ = }\log \left(\frac{1-0}{1+0}\right)\text{ = }\log 1\text{ = }0$,

$f\text{(1/2) = }\log \left(\frac{1-1/2}{1+1/2}\right)\text{ = }\log \frac{1}{3}etc$

ALGEBRA OF FUNCTIONS

Let two functions be f: D₁ $\to$ R and g: D₂ $\to$ R. We describe functions f + g, f - g, f. g and f /g as follows:

$\bullet$f + g : D $\to$R is a function defined by

(f + g)x = f(x) + g(x) where D = D₁$\cap$D₂

$\bullet$f - g : D$\to$ R is a function defined by

(f - g)x = f(x) –g(x) where D = D₁$\cap$D₂

$\bullet$f . g : D $\to$ R is a function defined by

by (f . g)x = f(x). g(x) where D = D₁$\cap$D₂

$\bullet$f / g : D $\to$ R is a function defined by

(f /g)x = $\frac{f(x)}{g(x)}$ where D = D₁$\cap${x$\in$D₂: g(x) $\ne$0}

Illustration 5 Let f(x) =$\text{si}{\text{n}}^{\text{-1}}\text{x}$., g(x) = $\sqrt{\left(3-x\right)\left(x+1\right)}$. Find f + g, f - g, f . g and f /g

Solution: (f+g)x = ${\sin }^{-1}x+\sqrt{\left(3-x\right)\left(1+x\right)}$, defined on -1$⩽$ x$⩽$1

(f-g)x = ${\sin }^{-1}x-\sqrt{\left(3-x\right)\left(1+x\right)}$, defined on -1$⩽$ x$⩽$1

(f.g)x =${\sin }^{-1}x\times \sqrt{\left(3-x\right)\left(1+x\right)}$, defined on -1$⩽$ x$⩽$1

(f/g)x = $\frac{{\sin }^{-1}x}{\sqrt{\left(3-x\right)\left(x+1\right)}}$, defined on -1 < x $⩽$1

Domain and range of a function

The set X is called the domain of the function. $f$ and set Y is called the co-domain. The set of the image of all elements of X under the function $f$ is called the range of $f$ and is denoted by $f$(x). It is obvious that range could be a subset of co-domain as we may have few elements in co-domain which are not the images of any element of the set X.

Thus range of $f$ i.e. $f$(x) = {$f$(x): x $\in$ X}. Clearly $f$(x) $\subseteq$ Y

PROBLEM BASED ON DOMAIN OF A REAL VALUED FUNCTION

Illustration 6. Find the domain of $f\text{(x)=}\frac{{\text{x}}^{\text{2}}\text{+1}}{{\text{x}}^{\text{2}}\text{-4}}$

Concept: Denominator should be non-zero for any function

Solution: x² – 4 $\ne$ 0 $\Rightarrow$ x $\ne$ $\pm$ 2

Hence domain is R – {–2, 2}

Illustration 7. Find the domain of $f\text{(x) = }{\left(\frac{\text{x}}{\text{1 - }\left|\text{x}\right|}\right)}^{\text{1/2000}}$

Concept: Expression under even root (i.e. square root, fourth root, sixth root etc) should not be negative.

Solution: f(x) is defined when $\frac{\text{x}}{\text{1}-\left|\text{x}\right|}\ge \text{ 0}$ ….(1)

Case 1: x $\ge$0 $f\text{(x)=}\frac{\text{x}}{\text{1-x}}$

For domain $\frac{\text{x}}{1-\text{x}}\ge 0\Rightarrow \frac{\text{x}}{\text{x}-1}\le 0$

$x\in [0,1)$…….(2)

Case 2: x < 0 $f(\text{x})=\frac{\text{x}}{1+\text{x}}$

For domain $\frac{x}{1+x}\ge 0x\in \left(-\infty ,-1\right)\cup \left[0,\infty \right)$

Rejecting the values of x$\in [0,\infty \text{)}$because they don't satisfy the inequality

x < 0.

We get $x\in \left(-\infty ,-1\right)$……(3)

Taking union of (2) and (3)

Domain =$\left(-\infty ,-1\right)\cup [0,1)$

Problem Based On Range Of A Real Valued Function

Illustration 8. If $f\left(x\right)=\frac{x^2+2x+3}{x},$then find the range of f(x).

Key concept: If f(x) is in the form of $\frac{P\left(x\right)}{Q\left(x\right)}$, where p(x) and Q(x) are polynomial function of second degree we use the concept of quadratic equation.

Solution: Let $\frac{x^2+2x+3}{x}=y$

$\Rightarrow {\text{x}}^2+(2-y)\text{x}+3=0$

$\because \text{x}\in R\therefore D\ge 0\Rightarrow {(2-y)}^2-4\times 1\times 3\ge 0$

$\Rightarrow y\in \left(-\infty ,2-2\sqrt{3}\right]\cup \left[2+2\sqrt{3},\infty \right)$