Question #bc5a2

Question

799 views around the world

You can reuse this answer
Creative Commons License

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)#