How do you find #f(x)# and #g(x)# when #h(x)= (x+1)^2 -9(x+1)# and #h(x)= (fog)(x)#?

The answers are:
#f(x)= x^2-9x# and
#g(x)= x+1#
but how?

1 Answer
Dec 12, 2017

See below.

Explanation:

Note that

#(f @ g)(x) = f(g(x))#

Now if #f(x) = x^2-9x# and #g(x) = x+1# then

#f(g(x)) = g(x)^2-9g(x) = (x+1)^2-9(x+1)#

NOTE

There are infinite #f(x)# and #g(x)# such that #f(g(x)) = (x+1)^2-9(x+1)#

Considering

#f(x) = a x^2 + b x + c# and
#g(x) = d x+e#

we have

#f(g(x)) = ad^2x^2+(b d +2ade)x+ae^2+be+c# now comparing coeficients

#{(ae^2+be+c+8=0),(b d +2ade+7=0),(ad^2=1):}#

Solving now for #a,b,c# we have

#{(a=1/lambda^2),(b=-(7lambda+2mu)/lambda^2),(c->(mu^2+7lambda mu -8lambda^2)/lambda^2),(d=lambda),(e=mu):}#

which is a two parameter set of compatible values.