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Sagot :
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]
### 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]
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