Answer :

f: R → R, f(x) = 2x – 3

g: R → R, g(x) = x^{3} + 5

fog = f[g(x)]

⇒ fog = f (x^{3} + 5)

⇒ fog = 2(x^{3} + 5) – 3

⇒ fog = 2x^{3} + 10 – 3

⇒ fog = 2x^{3} + 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

⇒ 2a^{3} + 7 = 2b^{3} + 7

⇒ a^{3} = b^{3}

⇒ 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 = 2x^{3} + 7

⇒ 2x^{3} = 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

Rate this question :

Fill in theMathematics - Exemplar

Let f : [2, ∞) <sMathematics - Exemplar

Let f : N →Mathematics - Exemplar

Fill in theMathematics - Exemplar

Let f :R →<Mathematics - Exemplar

Let f : [0, 1] <sMathematics - Exemplar

Which of the follMathematics - Exemplar