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:

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

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

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

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