Question

Let $f:R\to R$ be defined by

$f\left(x\right)=\frac{x}{\sqrt{1}+x^2}.$ Then

$\left(fofof\right)\left(x\right)=$
.

Answer 1

Use the definition of composition of functions.

Given: $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

$\left(fofof\right)\left(x\right)=f\left[\begin{array}{cc}f\left(\begin{array}{cc}\frac{x}{\sqrt{1+x^2}}& \\ & \end{array}\right)& \\ & \end{array}\right]$

$[\because (hogof\left)\right(x)=h(g\left(f\right(x\left)\right)\left)\right]$

$=f\left[\begin{array}{ccc}\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}& & \\ & & \end{array}\right]=f\left(\begin{array}{cc}\frac{x}{\sqrt{1+2x^2}}& \\ & \end{array}\right)$


$=\frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac{x}{\sqrt{1+3x^2}}$