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

Show that the function f:R→Rf:R \to R is defined by f(x)=xx2+1f\left( x \right) = \dfrac{x}{{{x^2} + 1}}, ∀x∈R\forall x \in R is neither one-one nor onto. Also, if g:R→Rg:R \to R is defined as g(x)=2x−1g\left( x \right) = 2x - 1, find fog(x).fog\left( x \right).

Explanation

Solution

A function f:X→Yf:X \to Y is defined to be one-one, if the images of distinct elements of XX under ff are distinct, i.e., for every x1,x2∈X{x_1},{x_2} \in X, f(x1)=f(x2)f\left( {{x_1}} \right) = f\left( {{x_2}} \right) implies x1=x2{x_1} = {x_2}. Otherwise, f is not one-one. Also, a function f:X→Yf:X \to Y is onto only if the range of f=Yf = Y.

Complete step-by-step answer:
Given, f:R→Rf:R \to R is defined by f(x)=xx2+1f\left( x \right) = \dfrac{x}{{{x^2} + 1}}, ∀x∈R\forall x \in R
(1)\left( 1 \right) ff is not one-one: Let x1,x2∈R{x_1},{x_2} \in R (domain) and f(x1)=f(x2)f\left( {{x_1}} \right) = f\left( {{x_2}} \right)
⇒x1x12+1=x2x22+1\Rightarrow \dfrac{{{x_1}}}{{{x_1}^2 + 1}} = \dfrac{{{x_2}}}{{{x_2}^2 + 1}}
⇒x1(x22+1)=x2(x12+1)\Rightarrow {x_1}\left( {{x_2}^2 + 1} \right) = {x_2}\left( {{x_1}^2 + 1} \right)
⇒x1+x1⋅x22=x2+x2⋅x12\Rightarrow {x_1} + {x_1} \cdot {x_2}^2 = {x_2} + {x_2} \cdot {x_1}^2
⇒x1+x1⋅x22−x2−x2⋅x12=0\Rightarrow {x_1} + {x_1} \cdot {x_2}^2 - {x_2} - {x_2} \cdot {x_1}^2 = 0
⇒(x1−x2)−x1x2(x1−x2)=0\Rightarrow \left( {{x_1} - {x_2}} \right) - {x_1}{x_2}\left( {{x_1} - {x_2}} \right) = 0
⇒(x1−x2)(1−x1x2)=0\Rightarrow \left( {{x_1} - {x_2}} \right)\left( {1 - {x_1}{x_2}} \right) = 0
⇒x1=x2\Rightarrow {x_1} = {x_2} or x1x2=1{x_1}{x_2} = 1
We note that there are points , x1{x_1} and x2{x_2} with x1≠x2{x_1} \ne {x_2} and f(x1)=f(x2)f\left( {{x_1}} \right) = f\left( {{x_2}} \right). For instance, if we take x1=1{x_1} = 1 and x2=12{x_2} = \dfrac{1}{2}, then we have f(x1)=25f\left( {{x_1}} \right) = \dfrac{2}{5} and f(x2)=25f\left( {{x_2}} \right) = \dfrac{2}{5}, but 2≠122 \ne \dfrac{1}{2}. Hence ff is not one-one.
(2) ff is not onto: Let y∈Ry \in R(Co-domain)
f(x)=yf\left( x \right) = y
⇒x1+x2=y\Rightarrow \dfrac{x}{{1 + {x^2}}} = y
⇒y(1+x2)=x\Rightarrow y\left( {1 + {x^2}} \right) = x
⇒yx2−x+y=0\Rightarrow y{x^2} - x + y = 0
This is the quadratic equation in variable xx, whose solution can be find by the formula:
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here, a=y,b=−1,c=ya = y,b = - 1,c = y
∴x=−1±(1)2−4⋅y⋅y2y\therefore x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4 \cdot y \cdot y} }}{{2y}}
⇒x=−1±1−4y22y\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - 4{y^2}} }}{{2y}}
Since x∈Rx \in R, ∴1−4y2⩾0\therefore 1 - 4{y^2} \geqslant 0
⇒(1+2y)(1−2y)=0\Rightarrow \left( {1 + 2y} \right)\left( {1 - 2y} \right) = 0
⇒−12⩽y⩽12\Rightarrow \dfrac{{ - 1}}{2} \leqslant y \leqslant \dfrac{1}{2}
So Range(f)∈[−12,12]Range\left( f \right) \in \left[ {\dfrac{{ - 1}}{2},\dfrac{1}{2}} \right]
Since, Range(f)≠R(Co−domain)Range\left( f \right) \ne R\left( {Co - domain} \right)
∴f\therefore f is not onto.
Hence ff is neither one-one nor onto.
Given, f(x)=xx2+1f\left( x \right) = \dfrac{x}{{{x^2} + 1}} and g(x)=2x−1g\left( x \right) = 2x - 1
Now, fog(x)=f[g(x)]fog\left( x \right) = f\left[ {g\left( x \right)} \right]
⇒fog(x)=f(2x−1)\Rightarrow fog\left( x \right) = f\left( {2x - 1} \right)
⇒fog(x)=2x−1(2x−1)2+1\Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{{{\left( {2x - 1} \right)}^2} + 1}}
⇒fog(x)=2x−14x2+1−4x+1\Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{4{x^2} + 1 - 4x + 1}}
⇒fog(x)=2x−14x2−4x+2\Rightarrow fog\left( x \right) = \dfrac{{2x - 1}}{{4{x^2} - 4x + 2}}

The value of fog(x)is2x−14x2−4x+2fog\left( x \right) is \dfrac{{2x - 1}}{{4{x^2} - 4x + 2}}

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