g f

g f Figure . Clearly f and g ◦ f are one-to-one. But g is not one-to-one. Thus from the above diagram it shows that the statement is not true.

Sets, Relations and Functions Example . Let f, g : R → R be defined as f ( x ) = x −| x | and g ( x ) = x + | x | . Find f ◦ g . We know | x | = − x if x ≤ if x > So f ( x ) = x − ( − x ) if x ≤ x − x if x > Thus f ( x ) = x if x ≤ if x > Also g ( x ) = x + ( − x ) if x ≤ x + x if x > Thus g ( x ) = if x ≤ x if x > Let x ≤ .

Then ( f ◦ g )( x ) = f ( g ( x )) = f ( x ) = x. The last equality is taken because x ≤ whenever x ≤ . Let x > . Then ( f ◦ g )( x ) = f ( g ( x )) = f ( x ) = x.

Thus ( f ◦ g )( x ) = x for all x . . . Inverse of a Function Let there be a bijection f : X → Y as given in the Figure .

. c b a X f Y z Figure . If we look this function in a mirror, we get a function from Y to X . Let us call that function as g .

Then g is a function from Y to X defined by g ( x ) = b, g ( y ) = c, g ( z ) = a . This function g is an example for the inverse of f . Now we define the inverse of a function. .

Functions Definition . Let f : X → Y be a bijection. The function g : Y → X defined by g ( y ) = x if f ( x ) = y , is called the inverse of f and is denoted by f − . If a function f has an inverse, then we say that f is invertible .

There is a nice relationship between composition of functions and inverse. Let f : X → Y be a bijection and g : Y → X be its inverse. Then g ◦ f = I X and f ◦ g = I Y where I X and I Y are identity functions on X and Y respectively. Moreover, if f : X → Y and g : Y → X are functions such that g ◦ f = I X and f ◦ g = I Y , then both f and g are bijections and they are inverses to each other; that is f − = g and g − = f .

Using the discussions above, the terms invertible and inverse can be defined in some other way as follows: Definition . A function f : X → Y is said to be invertible if there exists a function g : Y → X such that g ◦ f = I X and f ◦ g = I Y where I X and I Y are identity functions on X and Y respectively. In this case, g is called the inverse of f and g is denoted by f − . We may use this concept to prove some functions are bijective.

If f is a bijection, then f − ( y ) is nothing but the pre-image of y under f . Let us note that the inverses are defined only for bijections. If f is not one-to-one , then there exists a and b such that a = b and f ( a ) = f ( b ) . Let this value be y .

Then we cannot define f − ( y ) because both a and b are pre-images of y under f , as f − cannot assume two different values for y . If f is not onto, then there will be a y in Y without a pre-image. In this case also we cannot assign any value to f − ( y ) . For example, if A = { , , , } and f = { ( , ) , ( , ) , ( , ) , ( , ) } .

Then the range of f is { , , , } ; the inverse of f is { ( , ) , ( , ) , ( , ) , ( , ) } . Working Rule to Find the Inverse of Functions from R to R : Let f : R → R be the given function. i. write y = f ( x ) ; ii.

write x in terms of y ; iii. write f − ( y ) = the expression in y . iv. replace y as x .

Example . If f : R → R is defined by f ( x ) = x − prove that f is a bijection and find its inverse. Method : One-to-one : Let f ( x ) = f ( y ) . Then x − = y − ; this implies that x = y .

That is, f ( x ) = f ( y ) implies that x = y. Thus f is one-to-one. Onto : Let y ∈ R . Let x = y + .

Then f ( x ) = ( y + ) − = y . Thus f is onto. This also can be proved by saying the following statement. The range of f is R (how?) which is equal to the co-domain and hence f is onto.

Inverse Let y = x − . Then y + = x and hence x = y + . Thus f − ( y ) = y + . By replacing y as x , we get f − ( x ) = x + .

Sets, Relations and Functions Method : Let y = x − . Then x = y + . Let g ( y ) = y + . Now ( g ◦ f )( x ) = g ( f ( x )) = g ( x − ) = ( x − ) + = x.

( f ◦ g )( y ) = f ( g ( y )) = f y + = y + − = y. Thus, g ◦ f = I X and f ◦ g = I Y This implies that f and g are bijections and inverses to each other. Hence f is a bijection and f − ( y ) = y + . Replacing y by x we get, f − ( x ) = x + .

. . Algebra of Functions A function whose co-domain is R or a subset of R is called a real valued function. We can discuss many more operations on functions if it is real valued.

Let f and g be two real valued functions. Can we define addition of f and g ? Naturally we expect the sum of two functions to be a function. The value of f + g at a point x should be related to the values of f and g at x .

So to define f + g at a point x , we must know both f ( x ) and g ( x ) . In other words x must be in the domain of f as well as in the domain of g . And the natural way of defining f + g at x is f ( x ) + g ( x ) . So if we impose a condition that the domains of f and g to be the same, then we can define f + g .

In the same way we can define subtraction, multiplication and many more algebraic operations available on the set R of the real numbers. Definition . Let X be any set. Let f and g be real valued functions defined on X .

Define, for all x ∈ X • ( f + g )( x ) = f ( x ) + g ( x ) . • ( f − g )( x ) = f ( x ) − g ( x ) . • ( fg )( x ) = f ( x ) g ( x ) . • f g ( x ) = f ( x ) g ( x ) , where g ( x ) = .

• ( cf )( x ) = cf ( x ) , where c is a real constant. • ( − f )( x ) = − f ( x ) . Note that the domain may be any set, not necessarily a set of numbers. For example if X is a set of students of a class, f and g functions representing the marks obtained by the students in two tests, then the function f + g represent the total marks of the students in the two tests.

It is easy to see that the operations addition, subtraction, multiplication and division defined above satisfy the following properties. • ( f + g ) + h = f + ( g + h ) • f + g = g + f • + f = f + , where is the zero function defined by ( x ) = for all x . • f + ( − f ) = ( − f ) + f = • f ( g + h ) = fg + fh • ( c + c ) f = c f + c f where c and c are real constants. We can list many more properties of these operations.

The proofs are simple; however let us prove only one to show a way in which these properties can be proved. Let us prove f ( g + h ) = fg + fh . To prove f ( g + h ) = fg + fh we have to prove that ( f ( g + h ))( x ) = ( fg + fh )( x ) for all x in the domain. .

Functions Theorem . : If f and g are real-valued functions, then f ( g + h ) = fg + fh . Proof. Let X be any set and f and g be real-valued functions defined on X .

Let x ∈ X . ( f ( g + h ))( x ) f ( x )( g + h )( x ) (by the definition of product) f ( x )[ g ( x ) + h ( x )] (by the definition of addition) f ( x ) g ( x ) + f ( x ) h ( x ) (by the distributivity of reals) ( fg )( x ) + ( fh )( x ) (by the definition of product) ( fg + fh )( x ) (by the definition of addition) Thus ( f ( g + h ))( x ) = ( fg + fh )( x ) for all x ∈ X ; hence f ( g + h ) = fg + fh . . .

Some Special Functions Now let us see some special functions. (i) The function f : R → R defined by f ( x ) = a x n + a x n − + a x n − + . . .

+ a n − x + a n , where a i are constants, is called a polynomial function . Since the right hand side of the equality defining the function is a polynomial, this function is called a polynomial function. (ii) The function f : R → R defined by f ( x ) = ax + b where a = and b are constants, is called a linear function . A function which is not linear is called a non-linear function .

Clearly a linear function is a polynomial function. The graph of this function is a straight line; a straight line is called a linear curve; so this function is called a linear function. (one may come across different definitions for linear functions in higher study of mathematics.) (iii) Let a be a non-negative constant. Consider the function f : R → R defined by f ( x ) = a x .

If a = , x = then the function becomes the zero function and if a = , then function f : R → R defined by f ( x ) = a x is the constant function f ( x ) = . [See, Figures . and . ].

– – – – y = x y = x y = e x e Figure . – – – – y = e –x y = e x Figure . When a > , the function f ( x ) = a x is called an exponential function . Moreover, any function having x in the “power” is called as an exponential function.

e is a special irrational number lies between and . We will study more about e in the subsequent chapters. (iv) Let a > be a constant. The function f : ( , ∞ ) → R defined by f ( x ) = log a x is called a logarithmic function .

In fact, the inverse of an exponential function f ( x ) = a x on a suitable domain is called a logarithmic function. [See, Figure . ]. (v) The real valued function f defined by f ( x ) = p ( x ) q ( x ) on a suitable domain, where p ( x ) and q ( x ) are polynomials, q ( x ) = , is called a rational function .

In fact, the domain of this function is the set obtained from R by removing the real numbers at which q ( x ) = . Sets, Relations and Functions – e log x log e x log x Figure . (vi) If f is a real valued function such that f ( x ) = , then the real valued function g defined by g ( x ) = f ( x ) on a suitable domain is called the reciprocal function of f . The domain of g is the set obtained from R by removing the real numbers at which f ( x ) = .

For example, the largest possible domain of f ( x ) = x − is R −{ } . Let us see two more categories of functions. Definition . A function f : R → R is said to be an odd function if f ( − x ) = − f ( x ) for all x ∈ R .

It is said to be an even function if f ( − x ) = f ( x ) for all x ∈ R . [See, Figures . and . ].

f (– x ) f ( x ) – x Figure . f ( x ) – x Figure . The function defined by f ( x ) = x, f ( x ) = x and f ( x ) = x + x are some examples for odd functions. The functions defined by f ( x ) = x , f ( x ) = , f ( x ) = x + x and f ( x ) = | x | are some examples for even functions.

Note that the function f ( x ) = x + x is neither even nor odd. We can prove the following results. (i) The sum of two odd functions is an odd function. (ii) The sum of two even functions is an even function (iii) The product of two odd functions is an even function.

(iv) The product of two even functions is an even function. (v) The product of an odd function and an even function is an odd function. (vi) The only function which is both odd and even function is the zero function. (vii) The product of a positive constant and an even function is an even function.

(viii) The product of a negative constant and an even function is also an even function. (ix) The product of a constant and an odd function is an odd function. (x) There are functions which are neither odd nor even. Let us prove one of the above properties.

The other properties can be proved similarly. . Functions Theorem . : The product of an odd function and an even function is an odd function.

Proof. Let f be an odd function and g be an even function. Let h = fg . Now h ( − x ) = ( fg )( − x ) = f ( − x ) g ( − x ) = − f ( x ) g ( x ) (as f is odd and g is even) = − h ( x ) Thus h is an odd function.

This shows that fg is an odd function. If one function is not odd then don’t think that the function is an even function. There are plenty of functions which are neither even nor odd.