Question

The function $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$ is
A. an increasing function
B. a decreasing function
C. an even function
D. None of these

Answer 1

Hint: To check a function increasing or decreasing, first order derivative test is used. Find the differentiation of the given function and equate it with zero to find the critical points. Use the concept that a function is increasing in the interval on which the first order derivative of the function is positive. Otherwise, it is decreasing.

Formula Used:
Quotient Rule: If $f\left( x \right)$ and $g\left( x \right)$ be two functions of $x$ then the differentiation of the function $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is $\dfrac{d}{{dx}}\left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} - f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}}}{{{{\left\{ {g\left( x \right)} \right\}}^2}}}$, provided $g\left( x \right) \ne 0$
$\dfrac{d}{{dx}}\left( {{e^{mx}}} \right) = m{e^{mx}}$
$\dfrac{d}{{dx}}\left( c \right) = 0$, where $c$ is a constant.

Complete step by step solution:
The given function is $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$
Differentiating the function with respect to $x$, we get
$f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) - \left( {{e^{2x}} - 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Now, $\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) - \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} - 0 = 2{e^{2x}}$
and $\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) + \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} + 0 = 2{e^{2x}}$
So, $f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\left( {2{e^{2x}}} \right) - \left( {{e^{2x}} - 1} \right)\left( {2{e^{2x}}} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Simplify the expression.
Take the term $\left( {2{e^{2x}}} \right)$ as common from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left\{ {\left( {{e^{2x}} + 1} \right) - \left( {{e^{2x}} - 1} \right)} \right\}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( {{e^{2x}} + 1 - {e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Cancel out the term ${e^{2x}}$ from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( 2 \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}} = \dfrac{{4{e^{2x}}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Value of every perfect square expression is always positive and an exponential function is also always positive.
So, clearly, $f'\left( x \right) > 0$ for all real values of $x$.
Thus, the given function is an increasing function.

Option ‘A’ is correct

Note: Many students get confused about the condition for a function to be increasing and decreasing. They should remember that a function is increasing if the first order derivative of the function is positive for all real values of $x$ but if the first order derivative of a function is negative for all real values of $x$ then the function is decreasing. A function is even if the value of the function remains same after replacement of $x$ by $\left( { - x} \right)$ i.e. $f\left( { - x} \right) = f\left( x \right)$ and a function is odd if $f\left( { - x} \right) = - f\left( x \right)$.