<s>[INST] If $(f+g)(x)=4$ and $(f\circ g)(x)=7-4x$ What is the sum of the possible values for $g(2)$? 
If $f$ and $g$ are polynomial functions so that $(f+g)(x)=4$ and
$(f\circ g)(x)=7-4x$, what is the sum of the possible values for
$g(2)$?
$1)2\qquad\qquad\qquad2)3\qquad\qquad\qquad3)-3\qquad\qquad\qquad4)-2$

To solve this problem First I realized that $f$ and $g$ should be linear polynomials in to have $(f\circ g)(x)=-4x+7$. Hence $f(x)=ax+b$ and $g(x)=cx+d$.
$$(f+g)(x)=(a+c)x+(b+d)=4$$
$$(f\circ g)(x)=(ac)x+(ad+b)=-4x+7$$
Hence, $$a+c=0$$ $$b+d=4$$ $$ac=-4$$ $$ad+b=7$$
Then got two cases:
First, $a=2\quad, c=-2,\quad d=3,\quad b=1\quad \Rightarrow g(x)=-2x+3\Rightarrow g(2)=-1$
Second, $a=-2\quad,c=2,\quad d=-1,\quad b=5\quad \Rightarrow g(x)=2x-1\Rightarrow g(2)=3$
So the first choice is correct.
I wonder can we solve this problem with other approaches? ( Looing for approaches with less calculations and more elegance! )
 [/INST] Using the condition that $f(x) + g(x) = 4$,  $f(x) = ax + b, \ g(x) = -ax + (4-b)$, which results in:
$$f(g(x)) = f(-ax + 4 - b) = a(-ax+4-b) + b = 7 - 4x$$
hence $-a^2x = -4x \implies a = \pm 2$ and $a(4-b)+b = 7 \implies b = 1, 5$. Thus $g(x) = -2x+3$ or $2x-1$, so the sum of the possible values of $g(2)$ is $-1 + 3 = 2$.
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