Question #43f3b
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"How can molarity be used as a conversion factor?"
f(g(x)=#4 (6x)/(x-1) +2= (24x)/(x-1) +2#
Given, f(g(x) = g(x), hence,
#(24x)/(x-1) +2= (6x)/(x-1)#
#(18x)/(x-1) +2=0#
18x +2(x-1)=0
20x -2=0
20x=2
#x=1/10#
Given: #f(x) = 4x+2# and #g(x)= 6x/(x-1)#
#f(g(x)) = 4g(x)+2#
Normally, we would substitute #6x/(x-1)# for #g(x)# and then simplify but let's wait a bit to do that.
Instead, let's set #4g(x)+2# equal to #g(x)#:
#4g(x)+2=g(x)#
Subtract #g(x)+2# from both sides:
#3g(x)= -2#
Now let's substitute #6x/(x-1)# for #g(x)#:
#3(6x/(x-1))= -2#
Multiply both sides by #(x-1)#:
#3(6x)= -2(x-1)#
Distribute through the ()s:
#18x= -2x+2#
Add 2x to both sides:
#20x= 2#
Divide both sides by 20:
#x= 1/10#
Check
#g(1/10) = 6(1/10)/(1/10-1)#
#g(1/10) = -2/3#
#f(-2/3) = 4(-2/3)+ 2= -2/3 #
#4(-2/3)+2 = -2/3#
#-8/3+6/2 = -2/3#
#-2/3 = -2/3#
This checks.
#f(x)=4x+2 rArr f(g(x))=4g(x)+2.#
But, given that, #f(g(x))=g(x) rArr 4g(x)+2=g(x) rArr 3g(x)=-2#
#because, g(x)=(6x)/(x-1), :. 3{(6x)/(x-1)}=-2#
# rArr 18x=-2(x-1)=2-2x rArr 18x+2x=2 rArr 20x=2 rArr x=2/20=1/10.#
