Sure! Let's analyze the two functions you've provided: $y = \tan(2x)$ and $y = \cot(4 - x^2)$.

c) $y = \tan(2x)$

1. Domain:
$\tan(2x) \text{ is undefined where } \cos(2x) = 0 \, \Rightarrow \, 2x = \frac{\pi}{2} + k\pi \, (k \in \mathbb{Z}) \, \Rightarrow \, x = \frac{\pi}{4} + \frac{k\pi}{2} \, (k \in \mathbb{Z}).$
Therefore, the domain of $y = \tan(2x)$ is:
$x \in \mathbb{R} \setminus \left\{ \frac{\pi}{4} + \frac{k\pi}{2} : k \in \mathbb{Z} \right\}.$

2. Range:
The range is all real numbers:
$y \in \mathbb{R}.$

3. Period:
The function $\tan(kx)$ has a period of $\frac{\pi}{k}$. Therefore, for $y = \tan(2x)$:
$\text{Period} = \frac{\pi}{2}.$

4. Graph:
The graph of $\tan(2x)$ will have vertical asymptotes at $x = \frac{\pi}{4} + \frac{k\pi}{2}$. The function will oscillate between $-\infty$ and $+\infty$ between each pair of asymptotes.

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d) $y = \cot(4 - x^2)$

1. Domain:
The cotangent function is undefined wherever $\sin(4 - x^2) = 0$, which happens when:
$4 - x^2 = n\pi \, (n \in \mathbb{Z}) \Rightarrow x^2 = 4 - n\pi \Rightarrow x = \pm\sqrt{4 - n\pi}.$
Therefore, the domain consists of all real numbers except those points, and it depends on the values of $n$.

2. Range:
The range is all real numbers, similar to the tangent function.
$y \in \mathbb{R}.$

3. Behavior:
Since $y = \cot(4 - x^2)$ has vertical asymptotes at points where $4 - x^2 = n\pi$, the function will have a pattern of oscillation between these asymptotes.

4. Graph:
The graph of $y = \cot(4 - x^2)$ will have a series of oscillations, with each segment being similar to the cotangent function.

 Summary:

- For $y = \tan(2x)$, the function is periodic with a period of $\frac{\pi}{2}$ and has vertical asymptotes at specific points based on $k\pi/2$.
- For $y = \cot(4 - x^2)$, it has vertical asymptotes determined by solving $4 - x^2 = n\pi$ and oscillates between those asymptotes.

If you need specific calculations, like derivatives or values at certain points, please let me know!