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A geometric series is a sum of terms where each term after the first is found by multiplying the previous term by a constant called the common ratio. In this exercise, the function \(f(x)\) is expressed as a power series that can be identified as a geometric series. To rewrite it in the form of a geometric series, we rearrange it to:

• \(f(x) = \sum_{n=0}^{\infty} (-1)^n \frac{(x-5)^n}{3^n}\)

This series has a constant common ratio of \(-\frac{x-5}{3}\). Recognizing the series as a geometric series is crucial because it directly helps in determining the convergence properties and evaluating the function at specific points. The beauty of a geometric series lies in its simplicity and the fact that it can sum to a finite value when it converges.

Convergence for a geometric series specifies whether the sum approaches a finite limit as the number of terms goes to infinity. For a geometric series with a common ratio \(r\), convergence occurs when the absolute value of \(r\) is less than 1:

• \(|r| < 1\)

In our case, the common ratio is \(-\frac{x-5}{3}\). The inequality for convergence thus becomes:

• \(|-\frac{x-5}{3}| < 1\)

Simplifying this inequality gives the following condition for \(x\):

• \(|x-5| < 3\)

This translates to the range \(2 < x < 8\), which establishes the interval within which our series converges. Recognizing this convergence criterion is essential when dealing with power series, as it enables us to know where the series is valid.

The domain of a function refers to all the possible input values (\(x\)-values) that can be plugged into the function without leading to undefined operations or non-convergence. In the context of a power series, the domain is determined by the interval where the series converges.

• In our given power series \(f(x) = \sum_{n=0}^{\infty} (-1)^n \frac{(x-5)^n}{3^n}\), we found earlier that convergence occurs when \(2 < x < 8\).

Therefore, the domain of the function is \((2, 8)\), meaning \(x\) must be greater than 2 and less than 8. Any value of \(x\) outside this interval will result in the series not converging to a finite sum, which is why determining the domain is crucial when working with power series.

Function evaluation is the process of finding the value of a function at a particular input. For power series, once the domain is known, evaluating the function within this domain can be straightforward using known formulas for geometric series.

• For \(f(3)\): Substitute \(x = 3\) into the general form. The common ratio becomes \(-\frac{2}{3}\). The formula for the sum of a geometric series \(\frac{1}{1-r}\) helps us find \(f(3) = 3\).
• For \(f(6)\): Substitute \(x = 6\). Now the common ratio is \(-\frac{1}{3}\). Again using the sum formula, we calculate \(f(6) = \frac{3}{4}\).

Evaluating functions at specific points allows us to see the behavior of the series within its domain and is key in understanding the nature of power series. It also shows how convenient it can be to handle geometric series with known formulas.