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

Let's consider the function \[ y = \tan\left(\frac{\pi x}{36}\right) \] on the closed interval \([0, 9]\). ### Step 1. Determine the Domain and Continuity The function \(\tan \theta\) is continuous within any interval that does not include an odd multiple of \(\frac{\pi}{2}\). For our function, the argument is \[ \theta = \frac{\pi x}{36}. \] As \(x\) varies from 0 to 9, the angle \(\theta\) varies from \[ \theta(0) = \frac{\pi \cdot 0}{36} = 0 \quad \text{to} \quad \theta(9) = \frac{\pi \cdot 9}{36} = \frac{\pi}{4}. \] Since \(\frac{\pi}{4}\) is less than \(\frac{\pi}{2}\), \(\tan \theta\) is continuous on the interval. ### Step 2. Check for Critical Points First, compute the derivative of the function. We have \[ y = \tan\left(\frac{\pi x}{36}\right). \] Using the chain rule, the derivative is \[ y' = \sec^2\left(\frac{\pi x}{36}\right) \cdot \frac{\pi}{36}. \] Since \(\sec^2\left(\frac{\pi x}{36}\right) > 0\) for all \(x\) in \([0,9]\) (because \(\frac{\pi x}{36}\) is in \([0, \frac{\pi}{4}]\)), we conclude that \[ y' > 0 \quad \text{for all} \quad x \in [0,9]. \] This means that the function is strictly increasing on the interval. ### Step 3. Evaluate at Endpoints Since the function is strictly increasing, the absolute minimum will occur at the left endpoint and the absolute maximum at the right endpoint. - At \(x = 0\): \[ y(0) = \tan\left( \frac{\pi \cdot 0}{36} \right) = \tan(0) = 0. \] - At \(x = 9\): \[ y(9) = \tan\left( \frac{\pi \cdot 9}{36} \right) = \tan\left(\frac{\pi}{4}\right) = 1. \] ### Conclusion The absolute minimum on \([0,9]\) is at \[ (x, y) = (0, 0), \] and the absolute maximum on \([0,9]\) is at \[ (x, y) = (9, 1). \]