Finding range of function
$$\ \ f(x)=(\tan^{-1}(x))^3+(\cot^{-1}(x))^3$$
I am trying to solve above using Inequality
We know that $\tan^{-1}(x),\cot^{-1}(x)$ is defined for $x\in(-\infty,\infty)$
Using Inequality
$\displaystyle \frac{(\tan^{-1}(x))^3+(\cot^{-1}(x))^3}{2}\geq \bigg(\frac{\tan^{-1}(x)+\cot^{-1}(x)}{2}\bigg)^3=\frac{\pi^3}{64}$
So we have $\displaystyle \bigg(\tan^{-1}(x)\bigg)^3+\bigg(\cot^{-1}(x)\bigg)^3 \geq \frac{\pi^3}{32}$
Equality hold when $\displaystyle x=1$
Also at $\displaystyle f(-\infty)= -\frac{\pi^3}{8}$
Also at $\displaystyle f(\infty)=\frac{\pi^3}{8}$
So range of function as $\displaystyle \bigg[\frac{\pi^3}{32},\frac{\pi^3}{8}\bigg)$
But in wolframalpha , it shows answer as $\displaystyle \bigg(-\frac{\pi^3}{8},-\frac{\pi^3}{32}\bigg]\cup \bigg[\frac{\pi^3}{32},\frac{\pi^3}{8}\bigg)$
Please have a look on that problem, Thanks
Best Answer
Let us note that for $x \in (-\infty, \infty)$, range of $\tan^{-1}x$ is $\Big( \dfrac{-\pi}{2},\dfrac{\pi}{2} \Big)$ and that of $\cot^{-1}x$ is $(0,\pi)$. Also $\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$. Setting $\tan^{-1}x=y$, $$f(y) = y^3 + \Big( \frac{\pi}{2} - y \Big)^3 = \frac{\pi^3}{8} + \frac{3\pi}{2}\Big(y^2 - \frac{\pi}{2}y \Big)$$ On completing the square, $$f(y) = \frac{\pi^3}{32} + \frac{3\pi}{2}\Big( y - \frac{\pi}{4}\Big)^2 $$ Graph of $f(y)$ is an upward opening parabola with vertex at $\dfrac{\pi}{4}$. Since $y \in \Big( \dfrac{-\pi}{2},\dfrac{\pi}{2} \Big)$ and $f(y)$ is continuous, it attains minima at $\Big( \dfrac{\pi}{4},\dfrac{\pi^3}{32} \Big)$. For $y \rightarrow \Big( \dfrac{-\pi}{2} \Big)^+$, maximum value of $f(y)$ approaches $\dfrac{7\pi^3}{8}$.
Thus range of $f(x)$ is $\Big[\dfrac{\pi^3}{32},\dfrac{7\pi^3}{8} \Big)$. It is never negative.
Note that $f(-\infty) = \Big(\dfrac{-\pi}{2}\Big)^3 + (\pi)^3 = \dfrac{7\pi^3}{8}$.