Question

If \[f(x) = \dfrac{{({x^2} - 1)}}{{({x^2} + 1)}}\] , for every real $x$ , then the minimum value of $f$
(1) does not exist because $f$ is unbounded
(2) is not attained even though $f$ is bounded
(3) is equal to $1$
(4) is equal to $ - 1$

Answer 1

Hint: A real number is a value of a continuous quantity that can represents a distance along a line. Real numbers can be defined as the union of both the rational number and irrational numbers. They can be both positive or negative. Real number is denoted by . If we get the minimum value then, after solving the function we find the denominator and take the denominator as zero. After that we get the value of the variable and put this and get the result.

Complete step-by-step solution:
From the given data we get \[f(x) = \dfrac{{({x^2} - 1)}}{{({x^2} + 1)}}\]
We add and subtract $1$ , in the numerator of the given equation, we get
\[f(x) = \dfrac{{{x^2} + 1 - 1 - 1}}{{({x^2} + 1)}}\]
\[ \Rightarrow f(x) = \dfrac{{{x^2} + 1 - 2}}{{({x^2} + 1)}}\]
We separate the variable and constant part of the numerator and denominator , we get
\[ \Rightarrow f(x) = 1 - \dfrac{2}{{({x^2} + 1)}}\]
Now we have to find the minimum value of $f$
$f$ takes minimum value when the $\dfrac{2}{{{x^2} + 1}}$ takes maximum value
i.e., ${x^2} + 1$ gives the minimum value
when ${x^2} + 1$ gives the minimum value then we get $x = 0$
Therefore , minimum value of $f$ $ = \dfrac{{0 - 1}}{{0 + 1}}$
$ = \dfrac{{ - 1}}{1}$
$ = - 1$
$\therefore $ min $f = - 1$
Option (4) is correct .

Note: If we get a fraction \[\dfrac{{ax + b}}{{3x}}\] , then we separate variable and constant part from the numerator and get $\dfrac{{ax}}{{3x}} + \dfrac{b}{{3x}}$ .
We also solve the given problem using another method , we differentiate the given function and take equal zero. After that we get the value of the variable and put in the function.
Differentiate the given function \[f(x) = \dfrac{{({x^2} - 1)}}{{({x^2} + 1)}}\]
$f'(x) = \dfrac{{2x({x^2} + 1) - 2x({x^2} - 1)}}{{{{({x^2} + 1)}^2}}}$
Now we take $f'(x) = 0$
$ \Rightarrow \dfrac{{2x({x^2} + 1) - 2x({x^2} - 1)}}{{{{({x^2} + 1)}^2}}} = 0$
$ \Rightarrow 2x({x^2} + 1) - 2x({x^2} - 1) = 0$
$ \Rightarrow 2x + 2x = 0$
$ \Rightarrow 4x = 0$
$ \Rightarrow x = 0$
Put this in the function and get the minimum value of $f$
$f(x) = \dfrac{{ - 1}}{1}$
$ \Rightarrow \min f(x) = - 1$