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Sagot :
Let's analyze the function [tex]\( f(x)=\left(\frac{1}{6}\right)^x+2 \)[/tex] to determine its domain and range.
### Finding the Domain
1. Definition of the function: The function [tex]\( f(x)=\left(\frac{1}{6}\right)^x + 2 \)[/tex] is defined for all values of [tex]\( x \)[/tex]. This is because there are no restrictions on the exponent [tex]\( x \)[/tex] when the base is [tex]\(\frac{1}{6}\)[/tex] (a positive fraction less than 1).
- No square roots or logarithms are involved that could restrict [tex]\( x \)[/tex].
2. Conclusion on domain: Therefore, the domain of the function is all real numbers.
[tex]\[ \text{Domain: } \{x \mid x \text{ is a real number} \} \][/tex]
### Finding the Range
1. Behavior of the exponential part: The term [tex]\(\left(\frac{1}{6}\right)^x\)[/tex]:
- As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( x \to \infty \)[/tex]), [tex]\(\left(\frac{1}{6}\right)^x \)[/tex] gets closer to 0 because [tex]\(\left(\frac{1}{6}\right)^x\)[/tex] is a fraction raised to a larger and larger positive power.
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]), [tex]\(\left(\frac{1}{6}\right)^x \)[/tex] grows larger without bound because [tex]\( \left(\frac{1}{6}\right)^{-x} \)[/tex] becomes [tex]\( 6^x \)[/tex].
2. Adding the constant 2:
- Since [tex]\( \left(\frac{1}{6}\right)^x \)[/tex] approaches 0 as [tex]\( x \to \infty \)[/tex], the smallest value of [tex]\( f(x) \)[/tex] is [tex]\( 2 \)[/tex] approached asymptotically but never actually reaching.
- As [tex]\( x \to -\infty \)[/tex], [tex]\( \left(\frac{1}{6}\right)^x \to \infty \)[/tex], meaning [tex]\( f(x) \)[/tex] increases boundlessly.
3. Conclusion on range: Therefore, the function [tex]\( f(x) \)[/tex] will have values that start just over 2 and increase without bound.
[tex]\[ \text{Range: } \{y \mid y > 2\} \][/tex]
### Matching the Given Choices
From the choices provided:
1. [tex]\(\left\{x \left\lvert\, x>-\frac{1}{6}\right.\right\}\)[/tex]; [tex]\(\{y \mid y>0\}\)[/tex] — Incorrect domain and incorrect range.
2. [tex]\(\left\{x \left\lvert\, x>\frac{1}{6}\right.\right\}\)[/tex]; [tex]\(\{y \mid y>2\}\)[/tex] — Incorrect domain, correct range.
3. [tex]\(\{x \mid x \text{ is a real number}\}\)[/tex]; [tex]\(\{y \mid y>2\}\)[/tex] — Correct domain, correct range.
4. [tex]\(\{x \mid x \text{ is a real number}\}\)[/tex]; [tex]\(\{y \mid y>-2\}\)[/tex] — Correct domain, incorrect range.
Given our found domain and range, the correct option is:
[tex]\[ \boxed{3} \][/tex]
### Finding the Domain
1. Definition of the function: The function [tex]\( f(x)=\left(\frac{1}{6}\right)^x + 2 \)[/tex] is defined for all values of [tex]\( x \)[/tex]. This is because there are no restrictions on the exponent [tex]\( x \)[/tex] when the base is [tex]\(\frac{1}{6}\)[/tex] (a positive fraction less than 1).
- No square roots or logarithms are involved that could restrict [tex]\( x \)[/tex].
2. Conclusion on domain: Therefore, the domain of the function is all real numbers.
[tex]\[ \text{Domain: } \{x \mid x \text{ is a real number} \} \][/tex]
### Finding the Range
1. Behavior of the exponential part: The term [tex]\(\left(\frac{1}{6}\right)^x\)[/tex]:
- As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( x \to \infty \)[/tex]), [tex]\(\left(\frac{1}{6}\right)^x \)[/tex] gets closer to 0 because [tex]\(\left(\frac{1}{6}\right)^x\)[/tex] is a fraction raised to a larger and larger positive power.
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]), [tex]\(\left(\frac{1}{6}\right)^x \)[/tex] grows larger without bound because [tex]\( \left(\frac{1}{6}\right)^{-x} \)[/tex] becomes [tex]\( 6^x \)[/tex].
2. Adding the constant 2:
- Since [tex]\( \left(\frac{1}{6}\right)^x \)[/tex] approaches 0 as [tex]\( x \to \infty \)[/tex], the smallest value of [tex]\( f(x) \)[/tex] is [tex]\( 2 \)[/tex] approached asymptotically but never actually reaching.
- As [tex]\( x \to -\infty \)[/tex], [tex]\( \left(\frac{1}{6}\right)^x \to \infty \)[/tex], meaning [tex]\( f(x) \)[/tex] increases boundlessly.
3. Conclusion on range: Therefore, the function [tex]\( f(x) \)[/tex] will have values that start just over 2 and increase without bound.
[tex]\[ \text{Range: } \{y \mid y > 2\} \][/tex]
### Matching the Given Choices
From the choices provided:
1. [tex]\(\left\{x \left\lvert\, x>-\frac{1}{6}\right.\right\}\)[/tex]; [tex]\(\{y \mid y>0\}\)[/tex] — Incorrect domain and incorrect range.
2. [tex]\(\left\{x \left\lvert\, x>\frac{1}{6}\right.\right\}\)[/tex]; [tex]\(\{y \mid y>2\}\)[/tex] — Incorrect domain, correct range.
3. [tex]\(\{x \mid x \text{ is a real number}\}\)[/tex]; [tex]\(\{y \mid y>2\}\)[/tex] — Correct domain, correct range.
4. [tex]\(\{x \mid x \text{ is a real number}\}\)[/tex]; [tex]\(\{y \mid y>-2\}\)[/tex] — Correct domain, incorrect range.
Given our found domain and range, the correct option is:
[tex]\[ \boxed{3} \][/tex]
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