If F(x) = 2x+1 And G(x) = (x-1)/2, What Is F(g(x))? | Socratic
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3 Answers
Deepak G. Aug 5, 2016#=x#
Explanation:
#f(g(x))=2(x-1)/2+1# #=x-1+1# #=x#
Answer link Tony B Aug 5, 2016#f(g(x))=x#
Explanation:
Suppose I write this another way
Let #u=(x-1)/2" "larr g(x)#
Given that #color(brown)(f(x)=2x+1)# then
#color(brown)(f(u)=2color(blue)(u)+1)#
But #color(blue)(u=(x-1)/2)#
So #color(brown)(f(u)=2[ color(blue)((x-1)/2) ]+1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#f(u)# is exactly the same process as #f(g(x))#
#f(g(x)) = cancel(2)[ (x-1)/(cancel(2))]+1" "larr" but -1+1=0"#
#f(g(x))=x#
Answer link Ratnaker Mehta Aug 5, 2016#fog : RRrarr RR : (fog)(x)=x#.
Explanation:
We note that the functions #f and g# are real functions, i.e.,
# f : RR rarr RR ; f(x)=2x+1, and, g : RR rarr RR ; g(x)=(x-1)/2#.
Let #D_f and R_f# denote the Domain and Range of the fun. #f#
resp.
Now, recall that the fun. #fog# will be defined iff #R_gsubD_f#........#(star)#, &,
in the event, the fun. #fog : RR rarr RR# is defined by,
#fog(x)=f(g(x)), x in D_g#
In case of our funs. #f and#, we have, #R_g=RR=D_f#, so cond.#(star)#
is satisfied, #fog : RRrarrRR# is available.
# AAx in RR, fog(x)=f(g(x))=f(y),...............[say, where, y=g(x)=(x-1)/2]#
#=2y+1............[as, f(x)=2x+1]#
#=2{(x-1)/2}+1.........[as, y=g(x)=(x-1)/2]#
#=x#.
Hence, #fog : RRrarr RR : (fog)(x)=x#.
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