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Write down the equations of the asymptotes of the functions below \( g(x)=3 \tan (2 x) \) where \( x \in\left[-90^{\circ} ; 90^{\circ}\right] \)

Ask by Pollard Sherman. in South Africa
Feb 20,2025

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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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The Deep Dive

The function \( g(x) = 3 \tan(2x) \) has vertical asymptotes where the tangent function is undefined, which occurs at \( 2x = \frac{\pi}{2} + k\pi \) for any integer \( k \). Therefore, solving for \( x \) gives the asymptotes at \( x = \frac{\pi}{4} + \frac{k\pi}{2} \). Within the interval \( \left[-90^{\circ}; 90^{\circ}\right] \) or \( \left[-\frac{\pi}{2}; \frac{\pi}{2}\right] \), we find \( x = \frac{\pi}{4} \) (which is \( 45^\circ \)) and \( x = -\frac{\pi}{4} \) (which is \( -45^\circ \)). These are our vertical asymptotes: - \( x = 45^\circ \) - \( x = -45^\circ \)

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