To find the domain of the function $f(x) = \left(\frac{1}{3}\right)^{x-2} + 1$, let's analyze the expression step-by-step:

1. Identify the Expression Type: The function is an exponential function, given by $\left(\frac{1}{3}\right)^{x-2} + 1$.

2. Determine Domain Restrictions:

   • Exponential functions of the form $a^{x}$, where $a > 0$, are defined for all real values of $x$.
   • Since $\frac{1}{3} > 0$, there are no restrictions on $x$ for the function $\left(\frac{1}{3}\right)^{x-2} + 1$.

3. Conclusion:

   • Therefore, the domain of $f(x) = \left(\frac{1}{3}\right)^{x-2} + 1$ is all real numbers.

Answer in Interval Notation:

The domain is: $(-\infty, \infty)$

Would you like further details on exponential functions or have any additional questions?

Related Questions:

1. What is the range of $f(x) = \left(\frac{1}{3}\right)^{x-2} + 1$?
2. How does changing the base affect the domain and range of an exponential function?
3. What is the effect of shifting an exponential function vertically?
4. How can you find the inverse of an exponential function like $f(x) = \left(\frac{1}{3}\right)^{x-2} + 1$?
5. How does the graph of $f(x) = \left(\frac{1}{3}\right)^{x-2} + 1$ differ from $g(x) = 3^{x-2} + 1$?

Tip: For exponential functions, the domain is usually unrestricted, but the range often depends on vertical shifts and the base of the exponent.