To find the asymptotes of the function \( f(x) = \tan(4x - \pi) \) over the interval \( x = 0 \) to \( x = \frac{\pi}{2} \), we need to determine where the argument of the tangent function, \( 4x - \pi \), equals \( \frac{(2k+1)\pi}{2} \) for any integer \( k \). This is because the tangent function has vertical asymptotes wherever its argument equals \( \frac{(2k+1)\pi}{2} \).
### Step-by-Step Solution:
1. Set Up the Equation for the Asymptotes:
[tex]\[
4x - \pi = \frac{(2k + 1) \pi}{2}
\][/tex]
2. Solve for \( x \) in terms of \( k \):
[tex]\[
4x = \frac{(2k + 1) \pi}{2} + \pi
\][/tex]
[tex]\[
4x = \frac{(2k + 1) \pi + 2 \pi}{2}
\][/tex]
[tex]\[
4x = \frac{(2k + 3) \pi}{2}
\][/tex]
[tex]\[
x = \frac{(2k + 3) \pi}{8}
\][/tex]
3. Determine Valid Values of \( k \):
We need the values of \( x \) to lie within the interval \( 0 \leq x \leq \frac{\pi}{2} \).
Evaluate \( x \) for various \( k \) to find valid values:
[tex]\[
x = \frac{(2k + 3) \pi}{8}
\][/tex]
- For \( k = 0 \):
[tex]\[
x = \frac{3 \pi}{8} \approx 0.39269908169872414
\][/tex]
- For \( k = 1 \):
[tex]\[
x = \frac{5 \pi}{8} \approx 0.98174
\][/tex]
This value lies outside the interval \( 0 \leq x \leq \frac{\pi}{2} \).
- For \( k = -1 \):
[tex]\[
x = \frac{\pi}{8} \approx 0.39269908169872414
\][/tex]
Since \( x = \frac{5\pi}{8} \) is outside of the interval, we discard this and only keep values within the specified interval.
4. Check for More Values:
- For \( k = -2, -3 \):
[tex]\[
x \leq 0, \text{ which are outside the interval.}
\][/tex]
Thus, the valid \( x \) values within \( 0 \leq x \leq \frac{\pi}{2} \) for the asymptotes of the given function are:
[tex]\[
x = 0.39269908169872414 \quad \text{and} \quad x = 1.1780972450961724
\][/tex]
These are the locations of the asymptotes for the function [tex]\( f(x) = \tan(4x - \pi) \)[/tex] in the interval from [tex]\( x = 0 \)[/tex] to [tex]\( x = \frac{\pi}{2} \)[/tex].
