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If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then (fof of \()(x)\) is
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The correct answer is:
\(\frac{\mathrm{x}}{\sqrt{1+3 \mathrm{x}^2}}\)
\(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\)
\((f o f)(x)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}\)
\((fofof)(x)=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}\)
\((f o f)(x)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}\)
\((fofof)(x)=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}\)
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