Maximum value of $y=\tan^{-1}x- \tan^{-1}\frac{x}{3}$

Question

Let's suppose $f(x)=\tan^{-1}x$ and $g(x)=- \tan^{-1}\frac{x}{3}$

I know that the range of $f(x)$ is $$-\frac{\pi }{2} < \tan^{-1}x < \frac{\pi }{2}$$

and the range of $g(x)$ is the same. I used desmos to simulate the graph of $y$ and found that the range of it is $-\frac{\pi}{6}\leq y\leq\frac{\pi}{6}$ which means the maximum value is $\frac{\pi}{6}$. I don't quite understand how to get this value though. Any hints?

So how do you find the maximum of a function using derivatives? — Thomas Andrews, May 08 '23 at 13:35
I understand my mistake now. I forgot all about the derivate of arctanx , so I just didn't notice that I could solve for the extreme values and go from there. — kauselis3000, May 08 '23 at 13:43

score 3 · Accepted Answer · answered May 08 '23 at 13:34

Let $$f(x)=\tan^{-1}x- \tan^{-1}\frac{x}{3}$$ $$\implies f'(x)=\frac{6-2x^2}{x^4+10x^2+9}$$ To find the extreme values, set $f'(x)=0$ $$\implies \frac{6-2x^2}{x^4+10x^2+9}=0$$ $$\implies x=\pm\sqrt3$$ We can clearly see that $f(\sqrt3)=\frac{\pi}{6}$ and $f(-\sqrt3)=\frac{-\pi}{6}$

So the maximum value of $\tan^{-1}x- \tan^{-1}\frac{x}{3}$ is $$\color{blue}{\frac{\pi}{6}} $$

Maximum value of $y=\tan^{-1}x- \tan^{-1}\frac{x}{3}$

1 Answers1