Question

The function $\dfrac{\left( {{e}^{2x}}-1 \right)}{\left( {{e}^{2x}}+1 \right)}$ is:
(1) Increasing
(2) Decreasing
(3) Even
(4) Strictly increasing

Answer 1

Hint:Here in this question we have been asked to evaluate if the given function $\dfrac{\left( {{e}^{2x}}-1 \right)}{\left( {{e}^{2x}}+1 \right)}$ is increasing or decreasing. To answer this question we will first check if the function is even or not and then differentiate the function to evaluate whether it is increasing or not.

Complete step-by-step solution:
Now considering from the question we have been asked to evaluate if the given function $\dfrac{\left( {{e}^{2x}}-1 \right)}{\left( {{e}^{2x}}+1 \right)}$ is increasing or decreasing.
From the basic concepts we can say that $f\left( x \right)$ is an increasing function if $f'\left( x \right)>0$ and $f\left( x \right)$ is a decreasing function if $f'\left( x \right)<0$ .
Let us suppose that $f\left( x \right)=\dfrac{\left( {{e}^{2x}}-1 \right)}{\left( {{e}^{2x}}+1 \right)}$ now we will evaluate if the function is even or odd.
$\begin{align}
  & f\left( -x \right)=\dfrac{\left( {{e}^{-2x}}-1 \right)}{\left( {{e}^{-2x}}+1 \right)} \\
 & \Rightarrow f\left( -x \right)=\dfrac{\left( \dfrac{1}{{{e}^{2x}}}-1 \right)}{\left( \dfrac{1}{{{e}^{2x}}}+1 \right)} \\
 & \Rightarrow f\left( -x \right)=\dfrac{\left( 1-{{e}^{2x}} \right)}{\left( 1+{{e}^{2x}} \right)}=-f\left( x \right) \\
\end{align}$
Since $f\left( -x \right)=-f\left( x \right)$ , we can say that $f\left( x \right)$ is an odd function.
Now we need to evaluate the differentiate of the function. By doing that we will have $f'\left( x \right)=\dfrac{\left( {{e}^{2x}}+1 \right)\left( 2{{e}^{2x}} \right)-\left( {{e}^{2x}}-1 \right)\left( 2{{e}^{2x}} \right)}{{{\left( {{e}^{2x}}+1 \right)}^{2}}}$ since the formula for differentiation of division is $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}}$ and we know that $\dfrac{d}{dx}{{e}^{ax}}=a{{e}^{ax}}$ .
Now by simplifying the differentiate of the function we will get $f'\left( x \right)=\dfrac{4{{e}^{2x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}}$ . Here $f'\left( x \right)>0$ for all real values of $x$ .
Hence we can say that the given function $f\left( x \right)$ is an increasing function for all real values of $x$ .
Hence we will mark the option “1” as correct.

Note:During the process of answering questions of this type we should be sure with the concepts and formulae that we are going to apply in between the steps. If someone had a misconception as $f'\left( x \right)<0$ implies that $f\left( x \right)$ is a decreasing function then they will end up having a wrong conclusion.