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If $f(x)=\frac{4 x+x^{4}}{1+4 x^{3}}$ and $g(x)=\operatorname{In}\left(\frac{1+x}{1-x}\right)$, then what is the
value of $f^{\circ} \mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)$ equal to?
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value of $f^{\circ} \mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)$ equal to?
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Verified Answer
The correct answer is:
1
$\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+\mathrm{x}^{4}}{1+4 \mathrm{x}^{3}}, \mathrm{~g}(\mathrm{x})=\ln \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
$g\left(\frac{e-1}{e+1}\right)=\ell n\left(\frac{1+\left(\frac{e-1}{e+1}\right)}{1-\left(\frac{e-1}{e+1}\right)}\right)=\ell n\left(\frac{e+1+e-1}{e+1-e+1}\right)$
$=\ln \left(\frac{2 \ell}{2}\right)=\ell$ n $\mathrm{e}=1$
fo $\mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)=\mathrm{f}(1)=\frac{4(1)+(1)^{4}}{1+4(1)^{3}}=\frac{4+1}{1+4}=\frac{5}{5}=1$
$g\left(\frac{e-1}{e+1}\right)=\ell n\left(\frac{1+\left(\frac{e-1}{e+1}\right)}{1-\left(\frac{e-1}{e+1}\right)}\right)=\ell n\left(\frac{e+1+e-1}{e+1-e+1}\right)$
$=\ln \left(\frac{2 \ell}{2}\right)=\ell$ n $\mathrm{e}=1$
fo $\mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)=\mathrm{f}(1)=\frac{4(1)+(1)^{4}}{1+4(1)^{3}}=\frac{4+1}{1+4}=\frac{5}{5}=1$
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