# Graphing the graph of y=f(x) for $\tan(f(x))=\frac{x}{1-x^2}$ given that $f(0)=\pi$

I would like to know what the graph of $$\tan(f(x))=\frac{x}{1-x^2}$$ given that $$f(0)=\pi$$ looks like.

My attemt

$$\tan(f(x))=\tan(f(x)+k\pi)$$ for some intiger $$k$$. It follows that $$f(x)=\arctan{\frac{x}{1-x^2}}-k\pi$$ . Using the initial condition we find$$k=-1$$ and thus $$f(x)=\arctan{\frac{x}{1-x^2}}+\pi$$ when I graphed this and checked with desmos this does not look like $$\tan(f(x))=\tan(f(x)+k\pi)$$ as the limits for $$x\to \infty$$ differ. In my case as $$x\to\infty$$, $$f(x)$$ tends to $$\pi$$ but in the solution it tends to $$2\pi$$.

My Graph  Could someone explain to me what I have done wrong?

Edit I should add that we were told that we may assume that $$f(x)$$ is continues at +/- 1. Maybe this was they way of telling is that we need to make it look continius?

• All the equations here contain only one variable, hence cannot be drawn in the plane. Did you mean to set $f(x)=y$? May 14, 2020 at 16:09
• i believe that is what was meant here May 14, 2020 at 16:11
• @Allawonder yes sorry, in the question we were told to graph $y=h(x)$ and in writing this on stack exchange I used $f(x)$ instead. In essence, I want to find the graph of $y=f(x)$. Sorry for the confusion, let me know if that makes it clearer May 14, 2020 at 16:14
• @MathsWizzard So you want to draw the curve $$\tan y=\frac{x}{1-x^2}$$? May 14, 2020 at 16:20
• @Allawonder yes given that $y=\pi$ at $x=0$ May 14, 2020 at 16:20

It is true, that for any $$x\in \mathbb{R}\setminus \{-1,1\}$$ there must exist a $$k\in \mathbb{Z}$$, such that $$f(x) = \arctan(\frac{x}{1-x^2})+k\pi$$, but we cannot simply assume that the same $$k$$ holds for all $$x$$.
If we have further information, such as the information that $$f$$ is continuous on $$(-\infty, -1)\cup (-1,1)\cup (1,\infty)$$, then we must have a fixed constant on each interval, meaning that $$f(x) = \begin{cases} \arctan(\frac{x}{1-x^2}) + k_1\pi &,\text{for } x\in (-\infty,-1) \\ \arctan(\frac{x}{1-x^2}) + k_2\pi &,\text{for } x\in (-1,1) \\ \arctan(\frac{x}{1-x^2}) + k_3\pi &,\text{for } x\in (1,\infty) \end{cases}$$ And as you have already mentioned the initial condition $$f(0)=\pi$$ implies that $$k_2=1$$.
Now if we want to extend $$f$$ to a function on the entire real line in a way, such that it is continuous, then we must have $$f(-1) = \lim_{x\:\uparrow \: -1}\arctan(\frac{x}{1-x^2}) +k_1 \pi = \lim_{x\:\downarrow \:-1} \arctan(\frac{x}{1-x^2}) + \pi$$ and $$f(1) = \lim_{x\:\uparrow \: 1}\arctan(\frac{x}{1-x^2}) + \pi = \lim_{x\:\downarrow \: 1} \arctan(\frac{x}{1-x^2}) + k_3\pi$$ where $$\lim_{x \: \uparrow a}$$ and $$\lim_{x \: \downarrow \: a}$$ denotes left and right limits respectively. Using that, $$\lim_{y\rightarrow -\infty}\arctan(y) = -\pi/2$$ and $$\lim_{y\rightarrow \infty} \arctan(y) = \pi/2$$ we find that $$f(-1) = \frac12 \pi$$ and $$f(1) = \frac{3}{2} \pi$$ and therefore $$k_1 = 0$$ and $$k_3 = 2$$, thus finally we get that
$$f(x) = \begin{cases} \arctan(\frac{x}{1-x^2}) &,\text{for } x\in (-\infty,-1) \\ \frac12 \pi &,\text{for } x=-1 \\ \arctan(\frac{x}{1-x^2}) + \pi &,\text{for } x\in (-1,1) \\ \frac32 \pi &,\text{for } x=1 \\ \arctan(\frac{x}{1-x^2}) + 2\pi &,\text{for } x\in (1,\infty) \end{cases}$$