Let f:mathbbR→ mathbbR be defined as f(x)=x-1 and g:R→ 1,-1→ mathbbR be defined as g(x)= x^2 x^2-1. Then the function fog is:

$fog(x)=f(g(x))$fog(x)=f(g(x)) $=f(\frac{{x}^{2}}{{x}^{2}+1})$=f(x2+1x2​)

$=\frac{{x}^{2}}{{x}^{2}+1}-1$=x2+1x2​−1 $=\frac{{x}^{2}-{x}^{2}-1}{{x}^{2}+1}$=x2+1x2−x2−1​ $=\frac{-1}{{x}^{2}+1}$=x2+1−1​

We know that, $0\leq {x}^{2}<\infty ,\forall x\in R$0≤x2<∞,∀x∈R

$\Rightarrow 1\leq {x}^{2}+1<\infty ,\forall x\in R$⇒1≤x2+1<∞,∀x∈R $\Rightarrow 1\geq \frac{1}{{x}^{2}+1}>0,\forall x\in R$⇒1≥x2+11​>0,∀x∈R $\Rightarrow -1\leq \frac{-1}{{x}^{2}+1}<0,\forall x\in R$⇒−1≤x2+1−1​<0,∀x∈R

So, range of $fog(x)$fog(x) is $[-1,0)\subset R$[−1,0)⊂R.

Hence, the function $fog(x)$fog(x) is into function and $fog(-x)=f(g(-x))=\frac{-1}{{(-x)}^{2}+1}=\frac{-1}{{x}^{2}+1}=f(g(x))$fog(−x)=f(g(−x))=(−x)2+1−1​=x2+1−1​=f(g(x))

$\therefore fog(x)$∴fog(x) is an even function. So, it is a many one function.

Hence, $fog(x)$fog(x) is neither one-one nor onto function.