Question-48
For what value of \(\displaystyle c\) is the function \(\displaystyle f\) strictly increasing in \(\displaystyle ( -\infty ,\infty )\)?
\[ \begin{equation*} f( x) =cx+\frac{1}{x^{2} +3} \end{equation*} \]
For \(\displaystyle f\) to be strictly increasing in \(\displaystyle \mathbb{R}\), \(\displaystyle f^{\prime }( x) >0\) for all \(\displaystyle x\in \mathbb{R}\).
\[ \begin{equation*} f^{\prime }( x) =c-\frac{2x}{\left( x^{2} +3\right)^{2}} \end{equation*} \]
\(\displaystyle f^{\prime }( x) >0\) for all \(\displaystyle x\in \mathbb{R}\) provided \(\displaystyle c >\frac{2x}{\left( x^{2} +3\right)^{2}}\) for all \(\displaystyle x\in \mathbb{R}\). We need to see if there is a possibility for this. Therefore, consider:
\[ \begin{equation*} g( x) =\frac{x}{\left( x^{2} +3\right)^{2}} \end{equation*} \]
We notice that \(\displaystyle g( x)\rightarrow 0\) as \(\displaystyle x\rightarrow \pm \infty\). This shows that the x-axis is an asymptote to \(\displaystyle g\). We also note that \(\displaystyle g( x) < 0\) for \(\displaystyle x< 0\) and \(\displaystyle g( x) >0\) for \(\displaystyle x >0\). Let us try to understand \(\displaystyle g\) better using its first derivative:
\[ \begin{equation*} \begin{aligned} g^{\prime }( x) & =\frac{\left( x^{2} +3\right)^{2} -4x^{2}\left( x^{2} +3\right)}{\left( x^{2} +3\right)^{4}}\\ & \\ & =\frac{3\left( 1-x^{2}\right)}{\left( x^{2} +3\right)^{3}} \end{aligned} \end{equation*} \]
We see that \(\displaystyle x=\pm 1\) correspond to critical points. We can also see that:
\[ \begin{equation*} g^{\prime }( x) \ \text{is} \ \begin{cases} < 0, & x< -1\\ =0, & x=-1\\ >0, & -1< x< 1\\ =0, & x=1\\ < 0, & x >1 \end{cases} \end{equation*} \]
This suggests that \(\displaystyle g\) decreases from \(\displaystyle -\infty\) to \(\displaystyle -1\), reaches a local minimum at \(\displaystyle x=-1\), then increases from \(\displaystyle -1\) to \(\displaystyle 1\), reaches a local maximum at \(\displaystyle x=1\) and then decreases from \(\displaystyle 1\) to \(\displaystyle \infty\). This along with the fact that the x-axis is an asymptote to \(\displaystyle g\) suggests that \(\displaystyle x=-1\) has to be a global minimum and \(\displaystyle x=1\) has to be a global maximum. We have reached the following conclusion:
\[ \begin{equation*} g( x) \geqslant g( 1) =\frac{1}{16} \end{equation*} \]
Plugging this back into \(\displaystyle f^{\prime }( x)\), we see that:
\[ \begin{equation*} c >\frac{1}{8} \Longrightarrow f^{\prime }( x) >0,\ \forall x\in \mathbb{R} \end{equation*} \]