Answer
The function \( g(x) = 3 \tan(2x) \) has vertical asymptotes at \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \). There are no horizontal asymptotes for this function.
Solution
To find the asymptotes of the function \( g(x) = 3 \tan(2x) \), we need to analyze the function step by step.
### Step 1: Identify Vertical Asymptotes
The vertical asymptotes of the tangent function occur where the function is undefined, which is at the points where the argument of the tangent function is equal to \( \frac{\pi}{2} + k\pi \) for any integer \( k \).
For \( g(x) = 3 \tan(2x) \), we set the argument \( 2x \) equal to \( \frac{\pi}{2} + k\pi \):
\[
2x = \frac{\pi}{2} + k\pi
\]
Solving for \( x \):
\[
x = \frac{\pi}{4} + \frac{k\pi}{2}
\]
### Step 2: Determine Values of \( k \)
Given the domain \( x \in [-90^\circ, 90^\circ] \) (which is equivalent to \( x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)), we can find the values of \( k \):
1. For \( k = -1 \):
\[
x = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \quad (\text{valid})
\]
2. For \( k = 0 \):
\[
x = \frac{\pi}{4} \quad (\text{valid})
\]
3. For \( k = 1 \):
\[
x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \quad (\text{not valid, exceeds } \frac{\pi}{2})
\]
Thus, the vertical asymptotes are at:
\[
x = -\frac{\pi}{4} \quad \text{and} \quad x = \frac{\pi}{4}
\]
### Step 3: Identify Horizontal Asymptotes
The function \( g(x) = 3 \tan(2x) \) does not have horizontal asymptotes because the tangent function approaches infinity as \( x \) approaches the vertical asymptotes. Therefore, there are no horizontal asymptotes for this function.
### Summary of Asymptotes
- **Vertical Asymptotes:** \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \)
- **Horizontal Asymptotes:** None
If you need further assistance or clarification, feel free to ask!
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