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!
Best Answer
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 $-1.275982661$.