Question

Show that the function $f:R \to R$ is defined by $f\left( x \right) = \dfrac{x}{{{x^2} + 1}}$, $\forall x \in R$ 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

Hint: A function $f:X \to Y$ is defined to be one-one, if the images of distinct elements of $X$ under $f$ are distinct, i.e., for every ${x_1},{x_2} \in X$, $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$ implies ${x_1} = {x_2}$. Otherwise, f is not one-one. Also, a function $f:X \to Y$ is onto only if the range of $f = Y$.

Complete step-by-step answer:
Given, $f:R \to R$ is defined by $f\left( x \right) = \dfrac{x}{{{x^2} + 1}}$, $\forall x \in R$
$\left( 1 \right)$$f$ is not one-one: Let ${x_1},{x_2} \in R$ (domain) and $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$
$ \Rightarrow \dfrac{{{x_1}}}{{{x_1}^2 + 1}} = \dfrac{{{x_2}}}{{{x_2}^2 + 1}}$
$ \Rightarrow {x_1}\left( {{x_2}^2 + 1} \right) = {x_2}\left( {{x_1}^2 + 1} \right)$
$ \Rightarrow {x_1} + {x_1} \cdot {x_2}^2 = {x_2} + {x_2} \cdot {x_1}^2$
$ \Rightarrow {x_1} + {x_1} \cdot {x_2}^2 - {x_2} - {x_2} \cdot {x_1}^2 = 0$
$ \Rightarrow \left( {{x_1} - {x_2}} \right) - {x_1}{x_2}\left( {{x_1} - {x_2}} \right) = 0$
$ \Rightarrow \left( {{x_1} - {x_2}} \right)\left( {1 - {x_1}{x_2}} \right) = 0$
$ \Rightarrow {x_1} = {x_2}$ or ${x_1}{x_2} = 1$
We note that there are points , ${x_1}$ and ${x_2}$ with ${x_1} \ne {x_2}$ and $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$. For instance, if we take ${x_1} = 1$ and ${x_2} = \dfrac{1}{2}$, then we have $f\left( {{x_1}} \right) = \dfrac{2}{5}$ and $f\left( {{x_2}} \right) = \dfrac{2}{5}$, but $2 \ne \dfrac{1}{2}$. Hence $f$ is not one-one.
(2) $f$ is not onto: Let $y \in R$(Co-domain)
$f\left( x \right) = y$
$ \Rightarrow \dfrac{x}{{1 + {x^2}}} = y$
$ \Rightarrow y\left( {1 + {x^2}} \right) = x$
$ \Rightarrow y{x^2} - x + y = 0$
This is the quadratic equation in variable $x$, whose solution can be find by the formula:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a = y,b = - 1,c = y$
$\therefore x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4 \cdot y \cdot y} }}{{2y}}$
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - 4{y^2}} }}{{2y}}$
Since $x \in R$, $\therefore 1 - 4{y^2} \geqslant 0$
$ \Rightarrow \left( {1 + 2y} \right)\left( {1 - 2y} \right) = 0$
$ \Rightarrow \dfrac{{ - 1}}{2} \leqslant y \leqslant \dfrac{1}{2}$
So $Range\left( f \right) \in \left[ {\dfrac{{ - 1}}{2},\dfrac{1}{2}} \right]$
Since, $Range\left( f \right) \ne R\left( {Co - domain} \right)$
$\therefore f$ is not onto.
Hence $f$ is neither one-one nor onto.
Given, $f\left( x \right) = \dfrac{x}{{{x^2} + 1}}$ and $g\left( x \right) = 2x - 1$
Now, $fog\left( x \right) = f\left[ {g\left( x \right)} \right]$
$ \Rightarrow fog\left( x \right) = f\left( {2x - 1} \right)$
$ \Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{{{\left( {2x - 1} \right)}^2} + 1}}$
$ \Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{4{x^2} + 1 - 4x + 1}}$
$ \Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{4{x^2} - 4x + 2}}$

The value of $fog\left( x \right) is \dfrac{{2x - 1}}{{4{x^2} - 4x + 2}}$

Note: Given two functions $f$ and $g$, then $fog\left( x \right) = f\left[ {g\left( x \right)} \right]$ is known as composite function because it is the composition of $f$ and $g$. Note that the domain of $fog$ is the set of all real numbers $x$.