#### notes

In order to understant the composition of functions we have taken three sets in front of us name them as A,B and C.

now if we take a function f which is moving from A to B and then beginning with the stage set B, I take another function g from B to C. If we take a random element , after applying the funtion f it becomes f(x). Now beginning with f(x), this act as the element for the new function g and gives us the final product as g(f(x)), in short the new function which moves from A to C is defined by gof.

So if to define the function we say if

f: A→B & g: B→C, then

gof: A→C defined by gof(x)= g(f(x)) ∀x∈A

Notes:

1) f: A→B and g: B→C then,

range of f should be the starting point of the domain for the next function g for gof to exist. Likewise for fog to exist, the range of g should be subset of domain of f.

2) In general fog ≠ gof

Example- Let f: R→R

`"f"(x)= x^2`

g: R→R

g(x)= 2x+1

find fog & gof.

Solution- fog(x)= f(g(x))

fog(x)= f(2x+1)= (2x+1)^2

`gof(x)= g(f(x))`

`gof(x)= g(x^2)= 2x^2+1`

Invertible functions- Invertible means the function will be one one or onto both, that means the function will be bijective.

This means if f: A→B, then there exists `"f"^-1:"B"→"A"`

So, if this is was f(x)= y then `x= "f"^-1(y)`

Example1- If A= {1,2,3}, B= {4,5,6,7} and f={(1,4), (2,5), (3,6)} is a function from A to B. Is it one- one function?

Yes, because every element in set A have different image.

Example2- If f is invertible which is defined as `"f"(x)= (3x-4)/5` then write `"f"^-1 (x).`

Solution- f(x)=y

`(3x-4)/5 = y`

3x-4 = 5y

`x= (5y+4)/3 = "f"^-1 (y)`

Theorem 1: If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof) = (hog)of.

Proof : We have

ho(gof) (x) = h(gof(x)) = h(g(f(x))), ∀x in X

and (hog) of (x) = hog(f (x)) = h(g(f(x))), ∀x in X.

Hence, ho(gof) = (hog)of.

Theorem 2: Let f : X → Y and g : Y → Z be two invertible functions. Then gof is also invertible with `(gof)^(–1) = f^(–1)og^(–1).`

Proof: To show that gof is invertible with `(gof)^(–1) = f^(–1)og^(–1)`,

it is enough to show that `(f^(–1)og^(–1))o(gof) = "I"_X and (gof)o(f^(–1)og^(–1)) = "I"_Z.`

Now, `(f^(–1)og^(–1))o(gof) = ((f^(–1)og^(–1)) og) of`, byTheorem 1

= `(f^(–1)o(g^(–1)og)) of`, by Theorem 1

= `(f^(–1) oI_Y) of`, by definition of `g^–1`

= `"I"_X. `

Similarly, it can be shown that `(gof)o(f ^(–1) og ^(–1)) = "I"_Z.`