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Let $\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R}$ be defined by $f(x)=3 x^2-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^2+1}$, then $g \circ f$ is
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Verified Answer
The correct answer is:
$\frac{3 x^2-5}{9 x^4-30 x^2+26}$
Given that $f(x)=3 x^2-5$
and $g(x)=\frac{x}{x^2+1}$
$g o t=g\{f(x)\}=g\left(3 x^2-5\right)$
$\begin{aligned} & =\frac{3 x^2-5}{\left(3 x^2-5\right)^2+1}=\frac{3 x^2-5}{9 x^4-30 x^2+25+1} \\ & =\frac{3 x^2-5}{9 x^4-30 x^2+26}\end{aligned}$
and $g(x)=\frac{x}{x^2+1}$
$g o t=g\{f(x)\}=g\left(3 x^2-5\right)$
$\begin{aligned} & =\frac{3 x^2-5}{\left(3 x^2-5\right)^2+1}=\frac{3 x^2-5}{9 x^4-30 x^2+25+1} \\ & =\frac{3 x^2-5}{9 x^4-30 x^2+26}\end{aligned}$
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