To find the equation for fâg(x), plug the entire right side of the equation for g(x), for x in the equation for f(x). f(x) = 2x-1, g(x) = 3x+1 So we plug (3x+1) in place of x in 2x-1 and get 2(3x+1)-1 which simplifies to 6x+2-1 or 6x+1, so the equation for fâg(x) is fâg(x) = 6x+1. Now we must find the domain for fâg(x). Any value we substitute for x in fâg(x) must: (a) be in the domain of g(x). (b) g(x) must produce a value in the domain of f(x) g(x) has the domain ]-1; 2[ So the domain for fâg(x) must be contained entirely within this interval. Since g(x) is linear, the endpoints of the range of g(x) is found by substituting the end points of the domain of g(x) for x in the equation for g(x): 3(-1)+2=-1, 3(2)+2=8 so g(x) produces the range ]-1; 8[ But f(x) has the domain ]1; 10[ , so f(x) cannot be used for ]-1; 1], so we must remove ]-1; 1] from ]-1; 2[ which leaves ]1; 2[ So: fâg(x) = 6x+1, x â ]1; 2[ Now you switch the roles of f and g and do part b) yourself. Edwin