# Functions or Mappings

Let A B and be two non-empty sets. Then, a function f from set A to Bis a rule which associates elements of set A to elements of B such that all elements of set A are associated to elements of set B in unique way.

If f associates x $\in$ A  to y $\in$ B, then we say that y is the image of the element x and denote it by f(x) and write as y = f(x). The element x is called the pre-image or inverse image in B.

A function is denoted by f : A → B or A $\overset{f}{\rightarrow}$ B.

Note

• There may exist some elements in set B which are not the images of any element in set A.
• To each and every independent element in Athere corresponds one and only one image in B.
• Every function is a relation but every relation may or may not be a function.
• The number of functions from a finite set A into finite set B = $[n(B)]^{[n(A)]}$

Difference between a Relation and a Function

Geometrically, if we draw a vertical parallel line (VPL) i.e., any line which is parallel to y-axis (x = a), then if this line intersects the graph of the expression in more than one point, then the expression is a relation else if it intersects at only one point, the expression is a function, e.g.,

Domain, Codomain and Range of a Function

Let f : A → B, then A is known as domain of f while B is known as codomain of f.

Also, set f(A) = {f(x) : x $\in$ A} is known as range of f

Clearly, f(A) $\subseteq$ B

e.g., Let A = {1, –1, 2, –2}, B = {1, 4, 9}

f : A $\overset{x^2}{\rightarrow}$ B i.e f(x) = $x^2$

From the figure, it is clear that domain of function = {1, -1, 2, -2} and range of function = {1, 4}.

Also, codomain of function = {1, 4, 9}.

Sample Problem 1

Find the domain and range of the function f(x) = $\sqrt{9-x^2}$. [NCERT]
(a) [– 3, 3], [0, 3]
(b) [-3, 3), [0, 3]
(c) (– 3, 3), (0, 3)
(d) None of these

Given, f(x) = $\sqrt{9-x^2}$

For function to be defined,

9 - $x^2$ $\geqslant$ 0

⇒ $x^2$ - 9 $\leqslant$ 0             (*)

⇒ {x - (x - 3)} (x - 3) $\leqslant$ 0

⇒ -3 $\leqslant$ x $\leqslant$ 3

{if a < b and (x - a) (x - b) $\leqslant$ ⇒ a $\leqslant$ x $\leqslant$ b}

⇒ Domain = [-3, 3]

or Domain = {x: x $\in$ R and -3 $\leqslant$ x $\leqslant$ 3}

Again for range,

Let f(x) = y, then

⇒  y = $\sqrt{9-x^2}$

On squaring both sides, we get

$y^2$ = $\sqrt{9-x^2}$

⇒ $x^2$ = $\sqrt{9-y^2}$    (**)

On putting the value of $x^2$ in Eq. (*), we get

⇒ 9 - $y^2$ - 9 $\leqslant$ 0

⇒ -$y^2$ $\leqslant$ 0

⇒ $y^2$ $\geqslant$ 0

- $\infty$ < y < $\infty$

But y cannot be negative f(x) = $\sqrt{9-x^2}$ cannot contain negative values.

So, y $\geqslant$ 0

Also, from Eq. (**), x = $\sqrt{9 - y^2}$

⇒ 9 - $y^2$ $\geqslant$ 0

⇒ $y^2$ - 9 $\leqslant$ 0

-3 $\leqslant$ 0 y $\leqslant$ 3

⇒ Range = [0, 3] or {y:y $\in$ R, 0 $\leqslant$ y $\leqslant$ 3}

Sample Problem 2

Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from Z to Z defined by f(x) = ax + b for some integers a and b. Find, a, b.
(a) 1, 2
(b) 2, –1
(c) –1, –2
(d) None of these

Given, f = {(1, 1), (2, 3), (0, -1), (-1, -3)}

and f(x) = ax + b or y = ax + b

at x = 1, y = 1

⇒ 1 = a + b    (*)

at, x = 2, y = 3

⇒ 3 = 2a + b   (**)

On subtracting Eq. (*) from Eq. (**), we get a = 2

On putting the value of a in Eq. (*), we get

b = 1 - 2 ⇒ b = -1

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