If the function be defined by f(x) = 2x – 3 and by g(x) = x³ + 5 then find f o g and show that f o g is invertible. Also, find (fog)^-1, hence find (fog)^-1 (9).

f: R → R, f(x) = 2x – 3

g: R → R, g(x) = x³ + 5

fog = f[g(x)]

⇒ fog = f (x³ + 5)

⇒ fog = 2(x³ + 5) – 3

⇒ fog = 2x³ + 10 – 3

⇒ fog = 2x³ + 7

Now fog is invertible if fog is one to one and onto.

Let us check if fog is one to one

A function is said to be one to one if for every x there is at most one unique y

Or we can say that if fog(a) = fog(b) then a = b then fog is one to one

⇒ 2a³ + 7 = 2b³ + 7

⇒ a³ = b³

⇒ a = b

Hence fog is one to one

Now let’s check whether fog is onto

A function is said to be onto if for every y in co domain there exist a x in domain

fog is defined from R → R

We will find x in terms of y and if x ∈ R(domain) then fog is onto

⇒ y = 2x³ + 7

⇒ 2x³ = y – 7

Now for every y ∈ R, hence x belongs to domain hence fog is onto.

As fog is one to one and fog is onto hence fog is invertible.

fog^-1 means writing x in terms of y which we have already done in equation (a).

Hence

We have to find fog^-1(9)

⇒ fog^-1(9) = 1