Q. 245.0( 1 Vote )

If the function <

Answer :

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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