What are the domain and range of [tex]f(x)=\left(\frac{1}{6}\right)^x+2[/tex]?

A. Domain: [tex]\left\{x \mid x\ \textgreater \ -\frac{1}{6}\right\}[/tex]; Range: [tex]\{y \mid y\ \textgreater \ 0\}[/tex]
B. Domain: [tex]\left\{x \mid x\ \textgreater \ \frac{1}{6}\right\}[/tex]; Range: [tex]\{y \mid y\ \textgreater \ 2\}[/tex]
C. Domain: [tex]\{x \mid x \text{ is a real number}\}[/tex]; Range: [tex]\{y \mid y\ \textgreater \ 2\}[/tex]
D. Domain: [tex]\{x \mid x \text{ is a real number}\}[/tex]; Range: [tex]\{y \mid y\ \textgreater \ -2\}[/tex]



Answer :

To determine the domain and range of the function \( f(x) = \left(\frac{1}{6}\right)^x + 2 \), let's analyze each component step by step.

### Domain

1. Exponential Function Analysis:
- The function \( \left(\frac{1}{6}\right)^x \) is an exponential function with a base \( \left(\frac{1}{6}\right) \) which lies between 0 and 1. Exponential functions of the form \( a^x \), where \( 0 < a < 1 \), are defined for all real numbers \( x \).

2. Conclusion for Domain:
- There are no restrictions on \( x \) in the expression \( \left(\frac{1}{6}\right)^x \). Thus, the function \( f(x) \) is defined for all real numbers \( x \).

So, the domain of \( f(x) = \left(\frac{1}{6}\right)^x + 2 \) is:
[tex]\[ \{ x \mid x \text{ is a real number} \} \][/tex]

### Range

1. Original Exponential Function:
- Consider \( g(x) = \left(\frac{1}{6}\right)^x \). This function is always positive and approaches zero as \( x \) approaches positive infinity. As \( x \) decreases, \( \left(\frac{1}{6}\right)^x \) grows larger but still remains positive. Specifically, \( \left(\frac{1}{6}\right)^x > 0 \) for all real \( x \).

2. Adding a Constant:
- The given function \( f(x) \) adds 2 to the output of \( g(x) \). Therefore, for all \( x \):
[tex]\[ f(x) = \left(\frac{1}{6}\right)^x + 2 \][/tex]
Given \( \left(\frac{1}{6}\right)^x > 0 \), it follows that:
[tex]\[ f(x) = \left(\frac{1}{6}\right)^x + 2 > 0 + 2 \][/tex]
[tex]\[ f(x) > 2 \][/tex]

3. Conclusion for Range:
- The smallest value the function can approach is a value slightly greater than 2 (as \( \left(\frac{1}{6}\right)^x \) gets infinitesimally close to 0 but never reaches 0). Thus, \( f(x) > 2 \) for all \( x \).

So, the range of \( f(x) = \left(\frac{1}{6}\right)^x + 2 \) is:
[tex]\[ \{ y \mid y > 2 \} \][/tex]

### Correct Answer
Given the above analysis, the domain and range of the function \( f(x) = \left(\frac{1}{6}\right)^x + 2 \) are:

- Domain: \(\{ x \mid x \text{ is a real number} \}\)
- Range: \(\{ y \mid y > 2 \}\)

Therefore, the correct choice is:
[tex]\[ \{ x \mid x \text{ is a real number} \}; \text{ range: } \{ y \mid y > 2 \} \][/tex]