If #f(x) = 2x+1# and #g(x) = (x-1)/2#, what is #f(g(x))#?

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Answer 1 · 2016-08-05T18:16:53

#=x#

#f(g(x))=2(x-1)/2+1#
#=x-1+1#
#=x#

Answer 2 · 2016-08-05T18:27:43

#f(g(x))=x#

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 3 · 2016-08-05T18:50:52

#fog : RRrarr RR : (fog)(x)=x#.

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#.