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If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3 x-2$ and $g(x)=x^2+2$ then $[(g \circ f)+(f \circ g)](x)=$
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The correct answer is:
$12 g(x)-4 f(x)-22$
$f(x)=3 x-2$
$g(x)=x^2+2$
$\begin{aligned} & g \circ f(x)=g[f(x)]=f(x)^2+2 \\ & \& f \circ g(x)=f[g(x)]=3[g(x)]-2 \\ & (g \circ f+f \circ g)(x)=f(x)^2+2+3[g(x)]-2\end{aligned}$
$\begin{aligned} & =(3 x-2)^2+3\left(x^2+2\right) \\ & =9 x^2+4-12 x+3 x^2+6 \\ & =12 x^2-12 x+10 \\ & =12\left(x^2+2\right)-4(3 x-2)-22 \\ & =12 g(x)-4 f(x)-22\end{aligned}$
$g(x)=x^2+2$
$\begin{aligned} & g \circ f(x)=g[f(x)]=f(x)^2+2 \\ & \& f \circ g(x)=f[g(x)]=3[g(x)]-2 \\ & (g \circ f+f \circ g)(x)=f(x)^2+2+3[g(x)]-2\end{aligned}$
$\begin{aligned} & =(3 x-2)^2+3\left(x^2+2\right) \\ & =9 x^2+4-12 x+3 x^2+6 \\ & =12 x^2-12 x+10 \\ & =12\left(x^2+2\right)-4(3 x-2)-22 \\ & =12 g(x)-4 f(x)-22\end{aligned}$
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