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)
x3 = 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
x3 = 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) = x3 = 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
x3 = 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= y3
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)