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Answer 1 · 2017-11-15T21:31:13

$g(x)=sqrt(x) and g(x)=-sqrt(x)$

we take

$(gof)(x)=abs(x+1)$

$f(x)=x^2+2x+1=(x+1)^2$

now $f(x)=(x+1)^2$

for definition

$(gof)(x)=g(f(x))$

then

$abs(x+1)=g((x+1)^2)$

substitute $u=(x+1)^2$

isolate the variable x
$x=sqrt(u)-1$
now

$abs(x+1)=g(u)$

but when we have absolute valor, there are two cases, when the inside is positive or negative

when $abs(x+1)$ is negative is $-x-1$

$-x-1=g(u)$

but $x=sqrt(u)-1$

$-(sqrt(u)-1)-1=g(u)$
operating

$-(sqrt(u))=g(u)$

now replacing a random value for u we can replace any variable, returning to redefine this function in x

$-(sqrt(x))=g(x)$

similar to the first case now positive

$(x+1)=g(u)$

$x=sqrt(u)-1$

substitute

$sqrt(u)=g(u)$

now replacing a random value for u we can replace any variable, returning to redefine this function in x

$sqrt(x)=g(x)$