The strange "turning point" of $\arctan(x)/\arctan(\sqrt{x})$ After looking at an interesting graph:  $$y=\frac{\arctan(x)}{\arctan(\sqrt x)}$$
There seemed to be a turning point around $(3{,}88;1{,}198)$ (https://www.desmos.com/calculator/58wloddve3) <- A link to a graph of the equation
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When $x < 0$, there are no real solutions as the square root of a negative number will result in a complex number. (And we can leave complex numbers for another day!)
When $x > 0$ and $x < 1$, The graph makes sense, as $\frac {a}{b}$ where $b > a$, and $b \ne 0$ will always return a number greater than $0$ and less than $1$. (The graph also passes through $(1;1)$, which is obvious because $\frac {1}{1} = 1$)
However, when $x > 1$, The graph acts slightly strange.
$$\lim_{x\to \infty} \frac{\arctan(x)}{\arctan(\sqrt x)} = 1$$
But why is there a turning point around $(3{,}88;1{,}198)$? I look forward to any explanations and answers, thank you!
King Regards
Joshua
 A: I'm not sure what you really asking, but I'll try to illustrate the point.
For convenience, we will use another function:
$$f(x)=\frac{\arctan {x^2}}{\arctan {x}}$$
If you set $t=x^2$ you will get back to your initial function.
The plot of this function for $x>0$ looks like this:

There is an obvious maximum (it's not usually called a turning point). To find where this maximum is, we have to solve:
$$f'(x)=0$$
$$f'(x)=\frac{2x}{1+x^4} \frac{1}{\arctan x}-\frac{1}{1+x^2} \frac{\arctan x^2}{(\arctan x)^2}$$

$$\frac{2x(1+x^2)}{1+x^4} = \frac{\arctan x^2}{\arctan x}$$

Or we can rewrite this as:

$$x=\frac{1}{2}\frac{ (1+x^4)\arctan x^2 }{(1+x^2) \arctan x }$$

This equation can only be solved numerically (see @tired 's comment).
See the numerical solution by Wolfram Alpha.
$$x=\pm 1.96966420006958 \dots$$

$$x^2=3.87957706103574 \dots$$

As you can see, this fits well with the value you provided.

One way to solve this equation is Newton's method, which I'm not going to show explicitly for this case, since it will be quite cumbersome.
Another possible way is to use the series for $\arctan$ which work for the arguments with absolute value $<1$:
$$\arctan x=\frac{\pi}{2}-\arctan \frac{1}{x}=\frac{\pi}{2}-\left(\frac{1}{x}-\frac{1}{3x^3}+\frac{1}{5x^5}-\frac{1}{7x^7}+\dots \right)$$
Since we already have $x^4$ in our equation, it makes sense to leave all the terms with order $ \leq 4$, so we get an approximate equation:
$$\frac{2x(1+x^2)}{1+x^4} = \frac{\frac{\pi}{2}-\frac{1}{x^2}}{\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}}$$
It's actually a six degree polynomial equation, which has only one real positive root $x=1.95161\dots$, which is close enough to the root of the original equation.

Still, if you have a way to compute arctangents, I would recommend Newton's method.

