Q. 245.0( 1 Vote )

# If the function <

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

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