If fx axb f1x bxa with ab real what is the value of abSolut

If f(x)= ax+b f-1(x) = bx+a with a,b real, what is the value of a+b?

Solution

Let y= f(x)=ax+b. Then ax=y-b so that x=(y-b)/a. Hence, on interchanging x and y, we have f^-1(x) =(x-b)/a. However, since f^-1(x)=bx+a,we have bx+a=(x-b)/a or, bx+a = x/a –b/a. Now, on equating comparable terms, we have b = 1/a…(1) and a =-b/a…(2). On substituting b = 1/a in the 2^nd equation, we have a = 1/a(-1/a) or, a = -1/a².Now,on multiplying both the sides by a², we get a³= -1Further, since a is real, we have a = (-1)^1/3 = -1. Then from the 1^st equation, we have b = 1/a = 1/-1 = -1. Thus, a = -1 and b = -1.