338 views

Show that the function f, g :R ->R defined as :

$f(x)=x^{^{3}}$   ,  $g(x) = x^{\tfrac{1}{3}}$   x ∈ ℝ are inverses of each other.

a function can be inversed if it is bijective....

a function is bijective if it is one-one(injective) and onto(surjective)...

ALGORITHM  to check one- one

1) take two arbitrary elements x,y (say) in the domain of f...

2)put f(x)= f(y)

3) solve  f(x)= f(y) gives x=  y only..then f: A---->B is one- one

NOW...lets prove question is one- one

now let x,y be  arbitrary elements in the domain of f...such that

f(x) = f(y)

x= y3

x = y

hence ..it is one- one

now ALGORITHM  for ONTO(surjective)

let f:A--->B

1) choose any arbitrary elements y in B.

2) put f(x) = y

3) solve the equation f(x) = y for x and obtain  x in terms of y..let x= h(y)

4)if for all values of y∈B, the values of x obtained from x = h(y) are in A, then f is onto..if there are some y∈B for which x, given by x= g(y) is not in A..then f is not onto..

now prove above question is ONTO(surjective)..

f(x) = y

x= y

x= (y)1/3

clearly , for all y∈ R,  (y)1/3 is a real number

thus, for all y∈ R, B(co-domain) there exists x=  (y)1/3  in R, A(domain)  such that

f(x) = x= y

hence, f:R-->R is onto function..and hence above function is bijective

NOW , find inverse

ALGORITHM for INVERSE

let f: A-->B be a bijection ..

1) put f(x) = y

2)  solve the equation f(x) = y for x and obtain  x in terms of y

3) the relation obtained in above step-2 replace x by f-1(y) to obtain the requred  inverse of y..

4)later convert all in terms of x..by just changing y to x..

now lets solvee questtion..

f(x) =y

x= y

x =(y)1/3 put in place of x put it as  f-1(y)

f-1(y) =(y)1/3

now we can write it as: by changing y to x

f-1(x) =(x)1/3

now..we can clearly see..that  f-1(x) is equal to g(x) and hence f(x) is inverse of g(x)

now lets prove g(x) is inverse of

g(x) =y

y =(x)1/3

x= y

g-1(y) = y3 we can write it as

g-1(x) = x3 now we can clearly see that g-1(x) is equal to f(x)..g(x) is inverse of f(x)

1 comment

edited

@Tauhin. First you proved that 'g' is inverse of 'f'  and in the last argument, you proved the vice-versa.

For proving the inverse, you have used f(x) = y = x3

Thus, y$\frac{1}{3}$ = x = f-$-1$(y)

i.e. y$\frac{1}{3}$ = f-$-1$(y)

i.e. x$\frac{1}{3}$ = f-$-1$(x)

i.e. x$\frac{1}{3}$ = f-$-1$(x) = g(x)

i.e. f$-1$ = g

i.e 'g' is inverse of 'f'

and same line of reasoning for proving vice-versa.

fof-1(x) =x
let f-1(x)​ is inverse of f(x)

f(x) = x,  g(x) = x1/3
fog(x) = x                        //fog is R---> R

so f(x) is inverse of g(x).