Question

DRG of $\tan^{-1}\tan x^2$ when $x \in [-1,2]$

Answer 1

• Domain: $\{ x\in[-1,2] \mid x\neq \sqrt{\pi/2} \}$

• Range: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

• Graph:

  • For $|x|<\sqrt{\pi/2}$: $y=x^2$.
  • For $x\in(\sqrt{\pi/2},2]$: $y=x^2-\pi$,

  with a discontinuity (hole) at $x=\sqrt{\pi/2}$.

Answer 2

We wish to simplify

$$y=\tan^{-1}(\tan(x^2))$$

for $x\in[-1,2]$. Recall that $\tan^{-1}$ (or $\arctan$) returns the unique value in its principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Thus, for any real number $\theta$,

$$\tan^{-1}(\tan \theta)=\theta-k\pi,$$

where $k$ is an integer chosen so that

$$\theta-k\pi\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Since

$$\theta=x^2,$$

and noting that for $x\in[-1,2]$

$$x^2\in [0,4],$$

we separate the analysis into cases.

1. When is $x^2$ in the principal interval?

We need

$$x^2\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Since $x^2\ge0$, this becomes

$$x^2\in\left[0,\frac{\pi}{2}\right).$$

In this case, no adjustment is needed and

$$y=x^2.$$

2. When is $x^2$ outside the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$?

For $x^2\in \left[\frac{\pi}{2},4\right]$, we choose $k=1$ (since $x^2<3\pi/2$ for $x^2\le4$) so that

$$x^2-\pi\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Thus, for these $x$,

$$y=x^2-\pi.$$

3. Domain issues from the definition of $\tan$

Remember that the function $\tan(x^2)$ is not defined when

$$\cos(x^2)=0\quad\Longrightarrow\quad x^2=\frac{\pi}{2}+n\pi\quad (n\in\mathbb{Z}).$$

Within $x^2\in [0,4]$ the only problematic value is:

$$x^2=\frac{\pi}{2}\quad\Longrightarrow\quad x=\pm\sqrt{\frac{\pi}{2}}.$$

Since $x\in[-1,2]$ and $\sqrt{\frac{\pi}{2}}\approx1.253$, note that

$$-\sqrt{\frac{\pi}{2}}\approx -1.253 \notin [-1,2].$$

Thus the only point to be removed is

$$x=\sqrt{\frac{\pi}{2}}.$$

Final Answers

Domain:

$$\{ x\in[-1,2]: x\neq \sqrt{\pi/2}\}.$$

Range:

Looking at the two pieces:

• For $x^2\in[0,\frac{\pi}{2})$: $y=x^2$ covers $[0,\frac{\pi}{2})$.
• For $x^2\in\left(\frac{\pi}{2},4\right]$: $y=x^2-\pi$ covers values from just above $\frac{\pi}{2}-\pi=-\frac{\pi}{2}$ up to $4-\pi$.

The union of these intervals is

$$\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Graph:

The function is given piecewise by

• For $x$ such that $x^2<\frac{\pi}{2}$ (i.e. $|x|<\sqrt{\frac{\pi}{2}}$), $$y=x^2.$$
• For $x$ such that $x^2>\frac{\pi}{2}$ (i.e. for $x\in(\sqrt{\frac{\pi}{2}},2]$, and symmetrically for $x\in[-1, -\sqrt{\frac{\pi}{2}})$ if such existed in the interval—but note $-\sqrt{\frac{\pi}{2}} \notin[-1,2]$), $$y=x^2-\pi.$$

There is a jump discontinuity at $x=\sqrt{\frac{\pi}{2}}$ where the function is not defined.

A rough schematic using mermaid:

Minimal Core Explanation

1. For $x^2\in[0,\pi/2)$, $\tan^{-1}(\tan(x^2))=x^2$.
2. For $x^2\in(\pi/2, 4]$, choose $k=1$ so that $\tan^{-1}(\tan(x^2))=x^2-\pi$.
3. Exclude $x$ such that $x^2=\pi/2$ (i.e. $x=\sqrt{\pi/2}$) because $\tan$ is undefined there.
4. The combined range from the two cases is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.