Question

Find domain and range of \[f(x)=\dfrac{x}{1+{{x}^{2}}}\]
A. \[(-\infty ,-\infty )\]
B. \[(-1,1)\]
C. [\[\dfrac{-1}{2},\dfrac{1}{2}\]]
D. \[(-\sqrt{2},\sqrt{2})\]

Answer 1

Hint: Firstly we will find out the domain and apply the conditions of the domain by this we can find out the domain of the given function. Then to find out the range we will put \[f(x)\] is equal to \[y\] and then apply a quadratic formula and check which option is correct in the given options.

Complete step by step answer:
Domain of \[R\] contains only those elements of \[A\] which are related to \[B\] by the relation \[R\].Similarly the range of \[R\] consists of those elements of \[B\] which are related to \[A\] by the relation \[R\] .
A function is a relation and a relation may or may not be a function. Let \[y=f(x)\] be a function then the value of \[x\] approaches a quantity \[a\] after increasing from the left or decreasing from the right.
 Range is the set of all \[y\] values hence range is the set of all possible output values for a function in order from least to greatest. Domain is the set of all \[x\] values hence domain is the all possible input values for a function in order from least to greatest.

Now according to the question:
We have given the function:
\[\Rightarrow f(x)=\dfrac{x}{1+{{x}^{2}}}\]
This is an algebraic expression which is given in ratio form, hence we have to apply conditions for domain:
First condition is denominator must not be equal to zero
\[\Rightarrow 1+{{x}^{2}}\ne 0\]
Second condition is the value of \[{{x}^{2}}\] should not be equal to \[-1\]
\[\Rightarrow {{x}^{2}}\ne -1\]
By this we can conclude that:
\[x\in R\]
Hence domain will be \[x\in R\]
To find the range of \[f(x)=\dfrac{x}{1+{{x}^{2}}}\] , in this function in numerator there is a linear term and in denominator there is an quadratic expression:
\[\Rightarrow f(x)=y=\dfrac{x}{1+{{x}^{2}}}\]
Cross multiply the term:
\[y+y{{x}^{2}}=x\]
Now rearrange the obtained result we get:
\[y{{x}^{2}}-x+y=0\]
Now apply the quadratic formula and as we know that the solution of \[x\] are real hence
\[\Rightarrow D\ge 0\]
\[\Rightarrow {{b}^{2}}-4ac\ge 0\]
Where \[a=y,b=-1,c=y\]
Putting the values of \[a,b,c\] in the quadratic formula we get:
\[\Rightarrow {{(-1)}^{2}}-4\cdot y\cdot y\ge 0\]
\[\Rightarrow 1-4{{y}^{2}}\ge 0\]
Multiply the whole term by \[(-1)\] we will get:
\[\Rightarrow 4{{y}^{2}}-1\le 0\]
As we know that \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]
\[\Rightarrow (2y-1)(2y+1)\le 0\]
Critical points will be:
\[\Rightarrow y=[\dfrac{-1}{2},\dfrac{1}{2}]\]
Hence the range of the function will be \[[\dfrac{-1}{2},\dfrac{1}{2}]\]

So, the correct answer is “Option C”.

Note:
In algebra, the word function first appears in \[1673\] . Leibnitz and J. Bernoullie are the main contributors in the study of relations and functions. We must also remember that the inverse of a function is always unique and the inverse relation of functions is symmetric.