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If possible, find the values of 𝑎 ∈ 𝑅 for which the following function is continuous in its domain, if not possible, explain why

To determine the values of \( a \in \mathbb{R} \) that make the piecewise function continuous in its domain, we need to check the continuity of \( f(x) \) at the point where the pieces meet: at \( x = 9 \).

The given piecewise function is:

\[
f(x) = 
\begin{cases} 
\log_6\left( \frac{9 - x}{3 - \sqrt{x}} \right) & \text{if } x < 9 \\
a & \text{if } x = 9 \\
\frac{2}{\pi} \arctan\left( \frac{1}{x - 9} \right) & \text{if } x > 9 
\end{cases}
\]

### Continuity condition at \( x = 9 \):

A function is continuous at a point if:
1. The left-hand limit exists and is equal to the function value at the point.
2. The right-hand limit exists and is equal to the function value at the point.
3. The function value at the point is equal to the limit from both sides.

This means for continuity at \( x = 9 \), we need:
\[
\lim_{x \to 9^-} f(x) = f(9) = a = \lim_{x \to 9^+} f(x)
\]

#### Left-hand limit (\( \lim_{x \to 9^-} f(x) \)):

We need to compute the left-hand limit of the first piece:

\[
\lim_{x \to 9^-} \log_6\left( \frac{9 - x}{3 - \sqrt{x}} \right)
\]

As \( x \to 9^- \), the numerator \( 9 - x \to 0^+ \) and the denominator \( 3 - \sqrt{x} \to 0^+ \). So, this limit approaches:

\[
\lim_{x \to 9^-} \log_6\left( \frac{9 - x}{3 - \sqrt{x}} \right) = \log_6(1) = 0
\]

Thus, the left-hand limit is \( 0 \).

#### Right-hand limit (\( \lim_{x \to 9^+} f(x) \)):

We now compute the right-hand limit of the third piece:

\[
\lim_{x \to 9^+} \frac{2}{\pi} \arctan\left( \frac{1}{x - 9} \right)
\]

As \( x \to 9^+ \), \( \frac{1}{x - 9} \to +\infty \), and \( \arctan(\infty) = \frac{\pi}{2} \). So the limit becomes:

\[
\lim_{x \to 9^+} \frac{2}{\pi} \arctan\left( \frac{1}{x - 9} \right) = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1
\]

Thus, the right-hand limit is \( 1 \).

#### Continuity at \( x = 9 \):

For the function to be continuous at \( x = 9 \), we require that:

\[
\lim_{x \to 9^-} f(x) = \lim_{x \to 9^+} f(x) = a
\]

From the calculations above:
\[
0 = a = 1
\]

This is a contradiction, which means it is not possible to find a value of \( a \in \mathbb{R} \) such that the function is continuous at \( x = 9 \).

### Conclusion:
It is not possible to find a value of \( a \in \mathbb{R} \) that makes the function continuous at \( x = 9 \), because the left-hand and right-hand limits at \( x = 9 \) are different (0 and 1, respectively).

---

Would you like any more details or clarification on the solution?

Here are five follow-up questions you could explore:
1. What would happen if we change the function's definition for \( x > 9 \)?
2. How do we analyze differentiability for piecewise functions like this one?
3. Can we generalize this method to check continuity for more complicated piecewise functions?
4. What role does the logarithmic base play in the limit from the left?
5. What would change if the right-hand side used a different arctangent scaling factor?

**Tip**: When dealing with limits involving logarithmic or trigonometric functions, always double-check the approach to infinity or zero to ensure correct handling of asymptotic behavior.

If possible, find the values of 𝑎 ∈ 𝑅 for which the following function is continuous in its domain, if not possible, explain why.

Explore the continuity of a piecewise function with logarithmic and arctangent components, involving limits and evaluating continuity at a critical point. This problem is solved with detailed steps, including handling logarithmic limits and asymptotic behavior for inverse trigonometric functions.

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