To determine the range of the function [tex]\( f(x) = 3^x + 9 \)[/tex], let's analyze the behavior of the function:
1. Understand the base function [tex]\(3^x\)[/tex]:
- The function [tex]\(3^x\)[/tex] is an exponential function with base 3.
- Exponential functions of the form [tex]\(a^x\)[/tex] (where [tex]\(a > 1\)[/tex]) are strictly increasing.
- For any real number [tex]\(x\)[/tex], [tex]\(3^x\)[/tex] is always positive.
- When [tex]\(x = 0\)[/tex], [tex]\(3^x = 3^0 = 1\)[/tex].
- As [tex]\(x\)[/tex] increases, [tex]\(3^x\)[/tex] grows without bound (approaching [tex]\(\infty\)[/tex]).
- As [tex]\(x\)[/tex] decreases, [tex]\(3^x\)[/tex] approaches 0 but never reaches 0 (it gets arbitrarily close to 0).
2. Shift the function by adding 9:
- Adding 9 to [tex]\(3^x\)[/tex] vertically shifts the graph of [tex]\(3^x\)[/tex] upward by 9 units.
- When [tex]\(x = 0\)[/tex], [tex]\(f(x) = 3^x + 9 = 3^0 + 9 = 1 + 9 = 10\)[/tex].
- For any real number [tex]\(x\)[/tex], [tex]\(f(x) = 3^x + 9\)[/tex] is always greater than 9 because [tex]\(3^x\)[/tex] is always positive.
- As [tex]\(x\)[/tex] increases, [tex]\(f(x)\)[/tex] grows without bound (approaching [tex]\(\infty\)[/tex]).
- As [tex]\(x\)[/tex] decreases, [tex]\(f(x)\)[/tex] approaches 9 but never reaches 9 (it gets arbitrarily close to 9).
3. Determine the range:
- Since the function [tex]\(3^x\)[/tex] is always positive and strictly increasing:
- The smallest value of [tex]\(3^x\)[/tex] is 0 (never reached but approached).
- Thus, the smallest value of [tex]\(f(x) = 3^x + 9\)[/tex] is 9 (never reached but approached).
- Therefore, the function [tex]\(f(x)\)[/tex] takes on values greater than 9 and increases indefinitely.
Thus, the correct range of the function [tex]\( f(x) = 3^x + 9 \)[/tex] is:
[tex]\[
\boxed{\{y \mid y > 9\}}
\][/tex]
