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

Question

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

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Answer 1 · 2017-12-12T01:11:20

See below.

Note that

$(f \circ g) (x) = f (g (x))$ $(f @ g)(x) = f(g(x))$

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

$f (g (x)) = {g (x)}^{2} - 9 g (x) = {(x + 1)}^{2} - 9 (x + 1)$ $f(g(x)) = g(x)^2-9g(x) = (x+1)^2-9(x+1)$

NOTE

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

Considering

$f (x) = a x^{2} + b x + c$ $f(x) = a x^2 + b x + c$ and
$g (x) = d x + e$ $g(x) = d x+e$

we have

$f (g (x)) = a d^{2} x^{2} + (b d + 2 a d e) x + a e^{2} + b e + c$ $f(g(x)) = ad^2x^2+(b d +2ade)x+ae^2+be+c$ 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.