Question:
If $f(x)=\log \left(\frac{1+x}{1-x}\right)$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$, then $f(g(x))$ is equal to
(a) $f(3 x)$
(b) $\{f(x)\}^{3}$
(c) $3 f(x)$
(d) $-f(x)$
Solution:
(c) $3 f(x)$
$f(x)=\log \left(\frac{1+x}{1-x}\right)$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$
Now,
$\frac{1+g(x)}{1-g(x)}=\frac{1+\frac{3 x+x^{3}}{1+3 x^{2}}}{1-\frac{3 x+x^{3}}{1+3 x^{2}}}$
$=\frac{1+3 x^{2}+3 x+x^{3}}{1+3 x^{2}-3 x-x^{3}}$
$=\frac{(1+x)^{3}}{(1-x)^{3}}$
Then, $f(g(x))=\log \left(\frac{1+g(x)}{1-g(x)}\right)$
$=\log \left(\frac{1+x}{1-x}\right)^{3}$
$=3 \log \left(\frac{1+x}{1-x}\right)$
$=3 f(x))$