How do you find the compositions given #f(x) = 1/(1+x)# and #g(x) = sqrt(x+2)#?

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Answer 1 · 2016-01-22T10:37:49

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

#y=(g@f)(x)=sqrt(1/(1+x)+2)=sqrt((2x+3)/(1+x))#

Given
#f(x)= 1/(1+x)#

#g(x)=sqrt(x+2)#

To find #(f@g)(x)#, you can think:

#u=g(x)=sqrt(x+2)#

#:.y=(f@g)=f(u)=1/(1+u)=1/(1+sqrt(x+2))=#

#=1/(1+sqrt(x+2))*(1-sqrt(x+2))/(1-sqrt(x+2))=(1-sqrt(x+2))/(1-x-2)=#

#=-(1-sqrt(x+2))/(x+1)=(sqrt(x+2)-1)/(x+1)#

To find #(g@f)(x)#, you can think:

#u=f(x)=1/(1+x)#

#:.y=(g@f)=g(u)=sqrt(u+2)=sqrt(1/(1+x)+2)=#

#=sqrt((1+2+2x)/(1+x))=sqrt((2x+3)/(1+x))#