# Unfriendly function to study

Good morning/afternoon,

Our professor gave us this function to be studied:

$$y = f(x) = x + 2 - 3\arcsin\left(\frac{x^2-1}{x^2+1}\right)$$

But I am having many troubles with this.

Here is what I did:

Domain

This was rather easy for I needed to set the argument of the arcsine between $$-1$$ and $$1$$ and solve: $$D: \mathbb{R}$$.

Axis intersections

For $$x = 0$$, $$f(0) = 2 + \frac{3}{2}\pi$$ and that was ok. But here comes the pain: how can I solve the other intersection? $$y = 0$$ means some $$x$$ to be found... how?

Sign of the function

How to manage $$f(x) > 0$$ ?

Limits and asymptotes

That was easy (I hope): there are no vertical or horizontal asymptotes, but there is an obliquitous one: indeed

$$m = \lim_{x\to +\infty} \frac{f(x)}{x} = 1$$

$$q = \lim_{x\to +\infty} f(x) - mx = 2 + \frac{3}{2}\pi$$

Hence I have a line!

Max and min

Another trouble: Computing the derivative I got, explicitly

$$f'(x) = 1 - \frac{6x}{|x|(x^2+1)}$$

Which shows me that $$x = 0$$ is a non derivability point.

Plotting the function made me to see that $$x = 0$$ seems line a cusp point. But I did not understand why.

I tried to read the definition of a cusp (limits are infinite and of different signs, like in $$\sqrt{|x|}$$) but I cannot get why then $$1/x$$ has no cusp. Limits at $$0^+$$ and $$0^-$$ are infinite and of different signs!

Anyway: this put me on hold for I cannot go on with maxima and minima. I tried to solve it anyway, getting $$f'(x) = 0$$ with $$x = \pm\sqrt{5}$$ but it seems really wrong..

Any help? Thank you so much!

• On why 1/x has no cusp: note the function is not defined where it should have one! To get a cusp you need the derivatives to be different infinities on each side AND the function to be continuous there. Since the function is cts, you've found a cusp. – Artimis Fowl Mar 30 at 13:06

Since $$f(0) > 0$$ and $$f(x) \le x + 2 + 3 \pi/2 < 0$$ if $$x < -2 - 3 \pi/2$$, the Intermediate Value Theorem tells you $$f$$ switches from negative to positive somewhere between $$x=-2-3\pi/2$$ and $$x=0$$. The exact point where this occurs can only be determined numerically: it turns out to be approximately $$-1.275982661$$.
$$\frac{df}{dx}=1-6\frac{sgn(x)}{{1+x^2} }\implies \frac{df}{dx}=\frac{x^2+7}{x^2+1}, ~if~ x<0; \frac{df}{dx}=\frac{x^2-5}{x^2+1},~if~x>0$$ For $$x>0$$ its physical root is $$x=\sqrt{5}$$. Next $$f''(x)=12\frac{12x}{(x^2+1)^2}\implies f(+\sqrt{5})>0$$ So the function has a local minimum at $$x=\sqrt{5}.$$
The left derivative at $$x=0$$ is $$7$$ and the right one is $$-5$$. So $$f(x)$$ is non-differentiable at $$x=0$$ (there it is a corner not cusp)..
The function has a local maximum at $$x=0$$ of height equal to $$f(0)=\frac{4+3\pi}{2}$$