To determine which exponential function has an initial value of 2, let's evaluate each function at [tex]\( x = 0 \)[/tex].
Given the functions:
1. [tex]\( f(x) = 2 \cdot 3^x \)[/tex]
2. [tex]\( f(x) = 3 \cdot 2^x \)[/tex]
We can evaluate these functions at [tex]\( x = 0 \)[/tex]:
### Evaluating [tex]\( f(x) = 2 \cdot 3^x \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2 \cdot 3^0 \][/tex]
Since any number raised to the power of 0 is 1:
[tex]\[ f(0) = 2 \cdot 1 = 2 \][/tex]
So, for [tex]\( f(x) = 2 \cdot 3^x \)[/tex], the initial value when [tex]\( x = 0 \)[/tex] is 2.
### Evaluating [tex]\( f(x) = 3 \cdot 2^x \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 3 \cdot 2^0 \][/tex]
Similarly, any number raised to the power of 0 is 1:
[tex]\[ f(0) = 3 \cdot 1 = 3 \][/tex]
So, for [tex]\( f(x) = 3 \cdot 2^x \)[/tex], the initial value when [tex]\( x = 0 \)[/tex] is 3.
### Conclusion
The exponential function that has an initial value of 2 is [tex]\( f(x) = 2 \cdot 3^x \)[/tex].
