Question

If$f(x)=\log \left(\frac{1+x}{1-x}\right)$ and $g(x)=\frac{3x+x^3}{1+3x^2}$, then $f(g(x))$ is equal to

• A. $-f(x)$
• B. $3f(x)$
• C. $(f(x){)}^3$
• D. $f(3x)$

Answer 1

Answer: (B)

$f(g(x))=log\left(\frac{1+g(x)}{1-g(x)}\right)=\log \left(\frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}\right)$
$=\log \left(\frac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right)$
$=\log \left(\frac{1+x}{1-x}\right)=3\log \left(\frac{1+x}{1-x}\right)=3f(x)$.