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

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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

$$
(x, y) = (9, 1).
$$
The absolute minimum is at $$ (0, 0) $$ and the absolute maximum is at $$ (9, 1) $$.

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

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