Find the absolute extrema of the function on the closed interval. \[ y=\tan \left(\frac{\pi x}{36}\right),[0,9] \] minimum \( (x, y)=\left(\begin{array}{l}\text { maximum } \\ \text { ( } x, y)=\end{array}\right. \)

To find the absolute extrema of the function \( y=\tan \left(\frac{\pi x}{36}\right) \) on the closed interval \([0,9]\), we first evaluate the endpoints of the interval and any critical points within the interval. 1. **Endpoints:** - At \( x = 0 \): \[ y(0) = \tan(0) = 0 \] - At \( x = 9 \): \[ y(9) = \tan\left(\frac{9\pi}{36}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \] 2. **Finding Critical Points:** The function has vertical asymptotes where the tangent is undefined. The argument \( \frac{\pi x}{36} = \frac{\pi}{2} + k\pi \) leads to critical points: \[ x = 18 + 36k \] within the interval, only \( k=0 \): \[ x = 18 \quad \text{(not in the interval [0, 9])} \] Therefore, there are no critical points in the interval. 3. **Evaluating the values at the endpoints:** - Minimum value at \( (0, 0) \) - Maximum value at \( (9, 1) \) Therefore, the absolute minimum is at \( (0, 0) \) and the absolute maximum is at \( (9, 1) \). **Final Output:** minimum \( (x, y)=\left(0, 0\right) \) maximum \( (x, y)=\left(9, 1\right) \)