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

Question

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

Ariana Alvarez · Accepted Answer

See below.

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)# Taking into account #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):}# This is a set of compatible values for two parameters.

Ariana Alvarez · Answer

undefined To find $ f(x) $ and $ g(x) $ when $ h(x) = (f \circ g)(x) $, we need to express $ h(x) $ in terms of $ f(x) $ and $ g(x) $ and then solve for each function.

Given that $ h(x) = (x+1)^2 - 9(x+1) $ and $ h(x) = (f \circ g)(x) $, we can equate the two expressions:

$$ h(x) = f(g(x)) $$

Now, we can rewrite $ h(x) $ in terms of $ f(x) $ and $ g(x) $:

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

From the given expression, we can see that $ g(x) = x + 1 $ and $ f(x) = x^2 - 9x $.

Therefore, $ f(x) = x^2 - 9x $ and $ g(x) = x + 1 $.

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?

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