Question

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

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

Answer 1

Answer: (B)

$3f\left(x\right)$

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

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

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

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

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

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

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

$f\left(g\right(x\left)\right)=3f\left(x\right)$