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:


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!

  • 2
    $\begingroup$ 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. $\endgroup$ – 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 $-1.275982661$.

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$$\frac{df}{dx}=1-6\frac{sgn(x)}{{1+x^2} }\implies \frac{df}{dx}=\frac{x^2+7}{x^2+1}, ~if~ x<0; \frac{df}{dx}=\frac{x^2-5}{x^2+1},~if~x>0$$ 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 $f(0)=\frac{4+3\pi}{2}$

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