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

How to find the range of x21x2\dfrac{{{x^2}}}{{1 - {x^2}}} ?

Explanation

Solution

We are given an expression and asked to find the range of the given expression. The expression is in terms of xx so first try to find the value of xx. 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 x21x2\dfrac{{{x^2}}}{{1 - {x^2}}}.
We are asked to find the range of the given expression.
Let y=x21x2y = \dfrac{{{x^2}}}{{1 - {x^2}}} and try to solve for xx.
We have,
y=x21x2y = \dfrac{{{x^2}}}{{1 - {x^2}}}
y=1(1x2)1x2\Rightarrow y = \dfrac{{1 - \left( {1 - {x^2}} \right)}}{{1 - {x^2}}}
y=11x2(1x2)1x2\Rightarrow y = \dfrac{1}{{1 - {x^2}}} - \dfrac{{\left( {1 - {x^2}} \right)}}{{1 - {x^2}}}
y=11x21\Rightarrow y = \dfrac{1}{{1 - {x^2}}} - 1
y+1=11x2\Rightarrow y + 1 = \dfrac{1}{{1 - {x^2}}}
Multiplying both sides of the equation by 1x21 - {x^2} we get,
(1x2)(y+1)=(1x2)1x2\left( {1 - {x^2}} \right)\left( {y + 1} \right) = \dfrac{{\left( {1 - {x^2}} \right)}}{{1 - {x^2}}}
(1x2)(y+1)=1\Rightarrow \left( {1 - {x^2}} \right)\left( {y + 1} \right) = 1
Dividing both sides of the equation by (y+1)\left( {y + 1} \right) we get,
(1x2)(y+1)y+1=1y+1\dfrac{{\left( {1 - {x^2}} \right)\left( {y + 1} \right)}}{{y + 1}} = \dfrac{1}{{y + 1}}
(1x2)=1y+1\Rightarrow \left( {1 - {x^2}} \right) = \dfrac{1}{{y + 1}}
x2=1y+11\Rightarrow - {x^2} = \dfrac{1}{{y + 1}} - 1
x2=11(y+1)\Rightarrow {x^2} = 1 - \dfrac{1}{{\left( {y + 1} \right)}}
x2=y+11y+1\Rightarrow {x^2} = \dfrac{{y + 1 - 1}}{{y + 1}}
x2=yy+1\Rightarrow {x^2} = \dfrac{y}{{y + 1}}
For the solution to the real the term yy+1\dfrac{y}{{y + 1}} must be equal to or greater than zero.
That is,
yy+10\dfrac{y}{{y + 1}} \geqslant 0
For the above condition, there can be two situations.
1. y0y \geqslant 0 and y+1>0y + 1 > 0 which implies y[0,)y \in \left[ {0,\infty } \right).
2. y<0y < 0 and y+1<0y + 1 < 0 which implies y(,1)y \in \left( { - \infty , - 1} \right).

Therefore, the range of f(x)=x21x2f(x) = \dfrac{{{x^2}}}{{1 - {x^2}}} is y(,1)[0,)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.