Question

Show that the function $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1},\forall x\in R$ is neither one-one nor onto. Also, if $g:R\to R$ is defined as $g\left(x\right)=2x-1$, find $fog\left(x\right).$

Answer 1

For a function to be one to one, if we assume $f\left(x_1\right)=f\left(x_2\right),$ then $x_1=x_2$

Given, $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1},\forall x\in R$

Thus for one-one function, consider

$f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{x_1^2+1}=\frac{x_2}{x_2^2+1}$

$\Rightarrow x_1{x}_2(x_2-x_1)=x_2-x_1$

$\Rightarrow$ $x_2{x}_1=1,$ if $x_2\ne x_1$

$\Rightarrow f$ is not one-one function.

Also, a function is onto if and only if for every $y$ in the co-domain, there is $x$ in the domain such that $f\left(x\right)=y$

Let $f\left(x\right)=y$

$\Rightarrow \frac{x}{x^2+1}=y$

$\Rightarrow x=\frac{1\pm \sqrt{1-4y^2}}{2y}$

Now, substituting this $x$ in $f\left(x\right)=y$ we can see that this function is not onto.

Now, $fog\left(x\right)=f\left(g\right(x\left)\right)$

$=f(2x-1)$

$=\frac{2x-1}{4x^2-4x+2}$.