Question

The function \[f\left( x \right) = \dfrac{x}{{{x^2} + 1}}\] increasing, if
A. \[ - 1 < x\]
B. \[x > 1\]
C. \[ - 1 < x\] or \[x > 1\]
D. \[ - 1 < x < 1\]

Answer 1

Hint: We will first consider the given function. Then we will find the derivative of the given function with respect to \[x\] using the quotient rule. As we are given that the function is an increasing function which means the derivation of the function should be greater than zero. So, we will put the derivative greater than 0 and determine the interval in which the value of \[x\] lies.

Complete step by step answer:

We will first consider the given function, that is \[f\left( x \right) = \dfrac{x}{{{x^2} + 1}}\].
Now, we will find the derivative of the given function with respect to \[x\] using the quotient rule \[\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}\] where \[u = x,v = {x^2} + 1\]
We will first find \[\dfrac{{du}}{{dx}}\] and \[\dfrac{{dv}}{{dx}}\]
Thus, we get,
\[ \Rightarrow \dfrac{{du}}{{dx}} = 1\] and \[\dfrac{{dv}}{{dx}} = 2x\]
Now, we will substitute the values in the formula,
Thus, we have,
\[ \Rightarrow f'\left( x \right) = \dfrac{{\left( {{x^2} + 1} \right) \times 1 - x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\]
Now, we will simplify the right-hand side of the above expression,
\[
   \Rightarrow f'\left( x \right) = \dfrac{{{x^2} + 1 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \\
   \Rightarrow f'\left( x \right) = \dfrac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \\
 \]
Now, we are given in the question that the function is an increasing function, so, for the increasing function, its derivative is strictly greater than 0.
Therefore, we get,
\[
   \Rightarrow f'\left( x \right) > 0 \\
   \Rightarrow \dfrac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} > 0 \\
   \Rightarrow 1 - {x^2} > 0 \\
   \Rightarrow {x^2} < 1 \\
   \Rightarrow - 1 < x < 1 \\
 \]
Since as we do the square root, we get the two values one positive and other is negative.
Hence, we can conclude that \[x \in \left( { - 1,1} \right)\].
Thus, option D is correct.

Note: In the inequality, \[1 - {x^2} > 0\], the sign of inequality gets changes as the sign changes of \[x\] from negative to positive. For finding the derivative, we have used the quotient rule so remember the formula for it. Remember that for any increasing function, the derivative of that function will be greater than 0. Remember the quotient rule to find the derivative. Avoid doing calculation errors. Substitution should be done properly.