To determine the range of the function [tex]\( f(x) = \frac{1}{7} \cdot 9^x \)[/tex], we first need to understand its behavior as [tex]\( x \)[/tex] varies over all real numbers.
1. Exponentiation Component [tex]\( 9^x \)[/tex]:
- The expression [tex]\( 9^x \)[/tex] is an exponential function with a base greater than 1.
- For [tex]\( x \to -\infty \)[/tex], [tex]\( 9^x \)[/tex] approaches 0 but is always positive.
- For [tex]\( x \to \infty \)[/tex], [tex]\( 9^x \)[/tex] grows without bound.
2. Scaling Factor [tex]\( \frac{1}{7} \)[/tex]:
- Multiplying [tex]\( 9^x \)[/tex] by the constant [tex]\( \frac{1}{7} \)[/tex] scales the values but does not change the sign or the general positive nature of [tex]\( 9^x \)[/tex].
- Therefore, [tex]\( \frac{1}{7} \cdot 9^x \)[/tex] will be positive for all [tex]\( x \)[/tex].
3. Behavior Summary:
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to 0^+ \)[/tex] (approaches zero from the positive side).
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
Since [tex]\( f(x) = \frac{1}{7} \cdot 9^x \)[/tex] is always positive for any real number [tex]\( x \)[/tex] and can get arbitrarily close to zero but never actually reaches zero, and it can grow without bound, we conclude that the range is:
All real numbers greater than 0.
Therefore, the correct answer is:
[tex]\[ \text{all real numbers greater than 0} \][/tex]
