If f(x) = 1/(x+1) and g(x) = 2/(2x-1) how do you find g(f(x)) and its domain and range?

Question

Avery Alexander · Accepted Answer

#f(g(x)) = (2x-1)/(2x+1)#
with Domain of #RR-{-1/2}# and Range of #RR-{1}# It might help if we replace the variable #x# in #f(x)# with a different variable than the one used in #g(x)#. (We can do this because #x# is just an arbitrary place holder). Suppose for example we write: #color(white)("XXX")f(color(red)(w)) = 1/(color(red)(w)+1)# Then it might be easier to see how we can replace #color(white)("XXX")color(red)(w)color(white)("XXX")# with #color(white)("XXX")color(blue)(g(x))# #color(white)("XXX")f(color(blue)(g(x))) = 1/(color(blue)(g(x))+1)# #color(white)("XXXXXXX")=1/(color(blue)(2/(2x-1))+1)# #color(white)("XXXXXXX")=1/((2+2x-1)/(2x-1))# #color(white)("XXXXXXX")=(2x-1)/(2x+1)# The #f(g(x))#is defined for all Real values of #x# for which #color(white)("XXX")2x+1 !=0# #color(white)("XXX")x!=-1/2# That is, the Domain of #f(g(x))# is #RR-{-1/2}# #color(white)("XXX")#(or, if you prefer) #(-oo,-1/2)uu(-1/2,+oo)# One way to determine the range is to ask: "Is there any value, #c# for which #(2x-1)/(2x+1) = c# is impossible?" #color(white)("XXX")2x-1 = c(2x+1)# #color(white)("XXX")2x-2cx= c+1# #color(white)("XXX")x = (c+1)/(2(c-1))# This equation is clearly undefined if #c=1# Therefore the Range of #f(g(x))# is #color(white)("XXX")RR-{1}# #color(white)("XXX XX")# (...or, #(-oo,1)uu(1,+oo)#) This can also be seen from the graph of #(2x+1)/(2x-1)# graph{(2x-1)/(2x+1) [-5.546, 5.55, -2.773, 2.774]}

Kennedy Adams · Answer

undefined To find g(f(x)), first substitute f(x) into g(x), then determine the domain and range. So, g(f(x)) = 2 / (2 * (1 / (x + 1)) - 1). Simplifying this expression gives g(f(x)) = 2 / ((2 / (x + 1)) - 1), which further simplifies to g(f(x)) = 2 / ((2 - (x + 1)) / (x + 1)). Finally, g(f(x)) = 2 * (x + 1) / (2 - (x + 1)). The domain of g(f(x)) is all real numbers except x = -1, and the range is all real numbers except y = 2.