Mathematics COMPOSITE FUNCTION

Please Wait... While Loading Full Video

COMPOSITE FUNCTION

Composite Function

Let `f: A-> B` & `g: B -> C` be two functions. Then the fun ction `gof: A-> C` defined by

`(got) (x) ~ g(f(x)) AA x in A`

is called the composite ofthe two functions `f` & `g`. Diagramatically

`overset x (->) f overset (f(x)) (->) g -> g (f(x))`

Thus the image of every `x in A` under the function `gof` is the `g`- image of the `f`- image of `x`.

Note that gof is defined only if `AA x in A, f(x)` is an element of the domain of `g` so that we can take its `g`- image. Hence for `gof` of two functions `f` & `g`, the range off must be a subset of the domain of `g`.

Let `f: A-> B` & `g: B -> C` be two functions. Then the fun ction `gof: A-> C` defined by

`(got) (x) ~ g(f(x)) AA x in A`

is called the composite ofthe two functions `f` & `g`. Diagramatically

`overset x (->) f overset (f(x)) (->) g -> g (f(x))`

Thus the image of every `x in A` under the function `gof` is the `g`- image of the `f`- image of `x`.

Note that gof is defined only if `AA x in A, f(x)` is an element of the domain of `g` so that we can take its `g`- image. Hence for `gof` of two functions `f` & `g`, the range off must be a subset of the domain of `g`.

Properties Of Composite Functions

(i) The composite of functions is not commutative i.e. `gof != fog`.

(ii) The composite of functions is associative i.e. if `f, g, h` are three functions such that `fo(goh)` & `(fog)oh` are defined, then `fo(goh) = (fog)oh`.

Associatively: `f: (N) -> I_g f(x) = 2x`

`g: I_g -> Q \ \ \ \g(x) =1/x`

`h : Q-> R \ \ \ \ h(x) = e^(1/x)`

`(hog)of = ho(gof) =e^(2x)`

(iii) The composite of two bijections is a bijection i.e. if `f` and `g` are two bijections such that `gof` is defined, then `gof` is also a bijection.

Proof: Let `f: A-> B` and `g: B-> C` be two bijections. Then `gof` exists such that `gof: A-> C`

We have to prove that `gof` is one-one and onto.

One-one:

Let `a_1, a_2 in A` such that `(gof)(a_1)= (gof)(a_2)` , then

`(gof)(a_1) = (gof)(a_2) `

`=> g[f(a_1)] = g[f(a_2)]`

`=> f(a_1) = f(a_2) \ \ \ \ [ ∵ g text( is one- one)]`

`=> a_1 = a_2 \ \ \ \ [ ∵ f text( is one- one)]`

`:. gof` is also one-one function.

Onto: Let `c in C`, then

`c in C => 3 bin B` st `g (b) =c \ \ \ \ \ \ [ ∵ g text( is onto)]`

and `b in B => 3a in A ` s.t `f(a) =b \ \ \ \ \ \ [ ∵ f text( is onto)]`

Therefore, we see that

`c in C=>` {`3 a in A` S.t. `(got)(a) = g[f(a)] = g(b) = c`}

i.e. every element of `C` is the gof image of some element of `A`. As such `gof` is onto function. Hence `gof` befing one-one and onto, is a bijection.

(i) The composite of functions is not commutative i.e. `gof != fog`.

(ii) The composite of functions is associative i.e. if `f, g, h` are three functions such that `fo(goh)` & `(fog)oh` are defined, then `fo(goh) = (fog)oh`.

Associatively: `f: (N) -> I_g f(x) = 2x`

`g: I_g -> Q \ \ \ \g(x) =1/x`

`h : Q-> R \ \ \ \ h(x) = e^(1/x)`

`(hog)of = ho(gof) =e^(2x)`

(iii) The composite of two bijections is a bijection i.e. if `f` and `g` are two bijections such that `gof` is defined, then `gof` is also a bijection.

Proof: Let `f: A-> B` and `g: B-> C` be two bijections. Then `gof` exists such that `gof: A-> C`

We have to prove that `gof` is one-one and onto.

One-one:

Let `a_1, a_2 in A` such that `(gof)(a_1)= (gof)(a_2)` , then

`(gof)(a_1) = (gof)(a_2) `

`=> g[f(a_1)] = g[f(a_2)]`

`=> f(a_1) = f(a_2) \ \ \ \ [ ∵ g text( is one- one)]`

`=> a_1 = a_2 \ \ \ \ [ ∵ f text( is one- one)]`

`:. gof` is also one-one function.

Onto: Let `c in C`, then

`c in C => 3 bin B` st `g (b) =c \ \ \ \ \ \ [ ∵ g text( is onto)]`

and `b in B => 3a in A ` s.t `f(a) =b \ \ \ \ \ \ [ ∵ f text( is onto)]`

Therefore, we see that

`c in C=>` {`3 a in A` S.t. `(got)(a) = g[f(a)] = g(b) = c`}

i.e. every element of `C` is the gof image of some element of `A`. As such `gof` is onto function. Hence `gof` befing one-one and onto, is a bijection.

SiteLock