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If $f(x)=\frac{x}{2 x+1}$ and $g(x)=\frac{x}{x+1}$, then $(f \circ g)(x)=$
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Verified Answer
The correct answer is:
$\frac{x}{3 x+1}$
$$
\begin{aligned}
& (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\
& =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}
\end{aligned}
$$
$\begin{aligned} & (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\ & =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}\end{aligned}$
\begin{aligned}
& (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\
& =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}
\end{aligned}
$$
$\begin{aligned} & (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\ & =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}\end{aligned}$
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