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Georg FerdinandLudwig Philipp CantorBorn: 3 March 1845in St Petersburg,Russia Died: 6 Jan 1918 in Halle, Germany

GEORG CANTOR born in St petersburg, RUSSIA

TYPES OF FUNCTIONTYPES OF FUNCTION

One to One FunctionOne to One Function On to FunctionOn to Function One to One On to One to One On to Inverse of a FunctionInverse of a Function Equal FunctionEqual Function Identity FunctionIdentity Function Constant FunctionConstant Function Composite FunctionComposite Function

O N E T O O N E

A function f : A B is said to be

One to One Function. If no two distinct elements of A have the same image in B.

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On to FunctionOn to Function

f : A B is said to be an On to Function. If f(A) is the image of A f : A B is said to be an On to Function. If f(A) is the image of A equal B that is f is On to Function if every element of B. The Co-domain equal B that is f is On to Function if every element of B. The Co-domain is the image of at least one element A the domain.is the image of at least one element A the domain.

f: A B is on to for every x € B there exist at least onef: A B is on to for every x € B there exist at least one

x € A such that f(x) = y f(A) = B.x € A such that f(x) = y f(A) = B.

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A function f : A B is said to be a A function f : A B is said to be a bijection if it is both one to one and on to.bijection if it is both one to one and on to.

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INVERSE FUNCTIONINVERSE FUNCTION If f is a function then the set of If f is a function then the set of

ordered pairs obtained by ordered pairs obtained by interchanging the first and second interchanging the first and second coordinates of each order fair in F s coordinates of each order fair in F s called inverse of F. it denoted by Fcalled inverse of F. it denoted by F-1-1

f = { (0,0), (1,1), (2,4), (2,9)……..} f = { (0,0), (1,1), (2,4), (2,9)……..} f f -1-1 = { (0,0), (1,1), (4,2), (9,2) = { (0,0), (1,1), (4,2), (9,2)……..}……..}

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IDENTITY FUNCTION

A function f A→A is said to be an Identity Function on A denoted by IA .

f(x) = x

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CONSTANT FUNCTIONCONSTANT FUNCTION

A Function f : A→B is a constant A Function f : A→B is a constant function if there is an element cfunction if there is an element cЄЄB B such that f(x) =csuch that f(x) =c

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COMPOSITE FUNCTIONCOMPOSITE FUNCTION

Let F:A→B G:B→C be two functions Let F:A→B G:B→C be two functions then the composite function of F and then the composite function of F and G denoted by gof.G denoted by gof.

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f : A→B g : B→C

gof :A→C

GRAPHS OF FUNCTION

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Eg-1

Cuts the graph once

Eg-2 Eg-3Eg-4Eg-5Eg-6

Line l cuts the graph TWICE

Let f, g, h be functions defined as follows Let f, g, h be functions defined as follows f(x)=(x+2); g(x)=3x-1; h(x)= 2x show that f(x)=(x+2); g(x)=3x-1; h(x)= 2x show that

ho(gof)=(hog)ofho(gof)=(hog)of . . {ho[gof](x) {[hog]of}(x){ho[gof](x) {[hog]of}(x)={h(gof)(x)} =(hog)[f(x)]={h(gof)(x)} =(hog)[f(x)]=h{g[f(x)]} =h{g[f(x)]}=h{g[f(x)]} =h{g[f(x)]}=h[g(x+2)] =h[g(x+2)}=h[g(x+2)] =h[g(x+2)}=h[3(x+2)-1] =h[3(x+2)-1] =h[3(x+2)-1] =h[3(x+2)-1] =h(3x+5) =h(3x+5)=h(3x+5) =h(3x+5)=2(3x+5) =2(3x+5)=2(3x+5) =2(3x+5)=6x+10 =6x+10=6x+10 =6x+10 ho(go) = (hog)ofho(go) = (hog)of

Ex-1

EXERCISESEXERCISES

1.1. Sate and define types of functions.Sate and define types of functions.

2.2. Define Inverse of a function and Inverse Define Inverse of a function and Inverse function.function.

3.3. Let A={-1,1}. Let the functions fLet A={-1,1}. Let the functions f11 and f and f22

and f3 be from A into A defined as and f3 be from A into A defined as follows: ffollows: f11(x)=x; f(x)=x; f22(x)=x(x)=x22 ; f ; f33(x)=x(x)=x33..

4.4. Let f(x)=x2+2, g(x)=x2-2, for xLet f(x)=x2+2, g(x)=x2-2, for xЄЄR , find R , find fog(x), gof(x).fog(x), gof(x).

http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Cantor.html

Micro soft Encarta. Telugu Academy Text

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