Find the absolute extrema of the function on the closed interval. $$ f(x)=\cos (\pi x),\left[0, \frac{1}{6}\right] \text {, } \quad(x, y)=(\text { minimum }) $$ maximum $$ (x, y)=( $$ )

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1. Identify the function and the interval:  
   We have  
   $$
   f(x) = \cos (\pi x)
   $$  
   on the closed interval  
   $$
   \left[0, \frac{1}{6}\right].
   $$

2. Find the derivative:  
   Differentiate using the chain rule:  
   $$
   f'(x) = -\pi \sin (\pi x).
   $$

3. Find the critical points in the open interval by setting the derivative equal to $$0$$:  
   $$
   -\pi \sin (\pi x) = 0 \quad \Longrightarrow \quad \sin (\pi x) = 0.
   $$  
   The solutions to $$\sin (\pi x) = 0$$ are given by:  
   $$
   \pi x = k\pi \quad \Longrightarrow \quad x = k \quad \text{for any integer } k.
   $$  
   Within the interval $$\left(0, \frac{1}{6}\right)$$, the only possible value is $$x = 0$$, which is an endpoint.  
   There are no other interior critical points.

4. Evaluate the function at the endpoints:  
   - At $$x = 0$$:  
     $$
     f(0) = \cos (0) = 1.
     $$  
   - At $$x = \frac{1}{6}$$:  
     $$
     f\left(\frac{1}{6}\right) = \cos \left(\pi \cdot \frac{1}{6}\right) = \cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}.
     $$

5. Determine the absolute extrema:  
   - The absolute maximum value is $$1$$ at $$x = 0$$:  
     $$
     (x, y) = \left(0, 1\right).
     $$  
   - The absolute minimum value is $$\frac{\sqrt{3}}{2}$$ at $$x = \frac{1}{6}$$:  
     $$
     (x, y)=\left(\frac{1}{6}, \frac{\sqrt{3}}{2}\right).
     $$
The absolute maximum of $$ f(x) = \cos(\pi x) $$ on $$[0, \frac{1}{6}]$$ is $$1$$ at $$x = 0$$, and the absolute minimum is $$\frac{\sqrt{3}}{2}$$ at $$x = \frac{1}{6}$$.

Find the absolute extrema of the function on the closed interval.

maximum )

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Find the absolute extrema of the function on the closed interval. maximum )