Question

How to find the range of \[\dfrac{{{x^2}}}{{1 - {x^2}}}\] ?

Answer 1

Hint: We are given an expression and asked to find the range of the given expression. The expression is in terms of \[x\] so first try to find the value of \[x\]. Then observe the condition you obtain and check for the set of possible values. This will give you the range for the given expression.

Complete step by step solution:
Given, the expression \[\dfrac{{{x^2}}}{{1 - {x^2}}}\].
We are asked to find the range of the given expression.
Let \[y = \dfrac{{{x^2}}}{{1 - {x^2}}}\] and try to solve for \[x\].
We have,
\[y = \dfrac{{{x^2}}}{{1 - {x^2}}}\]
\[ \Rightarrow y = \dfrac{{1 - \left( {1 - {x^2}} \right)}}{{1 - {x^2}}}\]
\[ \Rightarrow y = \dfrac{1}{{1 - {x^2}}} - \dfrac{{\left( {1 - {x^2}} \right)}}{{1 - {x^2}}}\]
\[ \Rightarrow y = \dfrac{1}{{1 - {x^2}}} - 1\]
\[ \Rightarrow y + 1 = \dfrac{1}{{1 - {x^2}}}\]
Multiplying both sides of the equation by \[1 - {x^2}\] we get,
\[\left( {1 - {x^2}} \right)\left( {y + 1} \right) = \dfrac{{\left( {1 - {x^2}} \right)}}{{1 - {x^2}}}\]
\[ \Rightarrow \left( {1 - {x^2}} \right)\left( {y + 1} \right) = 1\]
Dividing both sides of the equation by \[\left( {y + 1} \right)\] we get,
\[\dfrac{{\left( {1 - {x^2}} \right)\left( {y + 1} \right)}}{{y + 1}} = \dfrac{1}{{y + 1}}\]
\[ \Rightarrow \left( {1 - {x^2}} \right) = \dfrac{1}{{y + 1}}\]
\[ \Rightarrow - {x^2} = \dfrac{1}{{y + 1}} - 1\]
\[ \Rightarrow {x^2} = 1 - \dfrac{1}{{\left( {y + 1} \right)}}\]
\[ \Rightarrow {x^2} = \dfrac{{y + 1 - 1}}{{y + 1}}\]
\[ \Rightarrow {x^2} = \dfrac{y}{{y + 1}}\]
For the solution to the real the term \[\dfrac{y}{{y + 1}}\] must be equal to or greater than zero.
That is,
\[\dfrac{y}{{y + 1}} \geqslant 0\]
For the above condition, there can be two situations.
1. \[y \geqslant 0\] and \[y + 1 > 0\] which implies \[y \in \left[ {0,\infty } \right)\].
2. \[y < 0\] and \[y + 1 < 0\] which implies \[y \in \left( { - \infty , - 1} \right)\].

Therefore, the range of \[f(x) = \dfrac{{{x^2}}}{{1 - {x^2}}}\] is \[y \in \left( { - \infty , - 1} \right) \cup \left[ {0,\infty } \right)\].

Note: Range of a function can be defined as the set of all possible values for the output of the function. There is another term called domain. Domain of a function is the set of all possible values for the input of the function. While finding the domain and range of a function, remember a few points. First, the denominator of the function cannot be zero as it will make the function infinite and secondly if there is a square root then the number under the square root must be a positive number as if we take the number to be negative the result will be a complex number.