To analyze the functions \( f(x) = 5x + 12 \) and \( g(x) = 3 \cdot 2^x \):
For small values of \( x \), the function with greater values is \( f(x) \). For instance, if we take \( x = 0 \):
- \( f(0) = 5(0) + 12 = 12 \)
- \( g(0) = 3 \cdot 2^0 = 3 \)
Thus, \( f(0) = 12 \) is greater than \( g(0) = 3 \).
Whereas, if we take \( x = 1 \):
- \( f(1) = 5(1) + 12 = 17 \)
- \( g(1) = 3 \cdot 2^1 = 6 \)
Again, \( f(1) = 17 \) is greater than \( g(1) = 6 \).
As the value of \( x \) increases, however, \( g(x) \) eventually has greater values. For instance, if we consider \( x = 5 \):
- \( f(5) = 5(5) + 12 = 37 \)
- \( g(5) = 3 \cdot 2^5 = 3 \cdot 32 = 96 \)
In this case, \( g(5) = 96 \) is greater than \( f(5) = 37 \).
As \( x \) continues to increase, \( g(x) \) grows exponentially due to its exponential component, while \( f(x) \) grows linearly. Therefore, for sufficiently large \( x \), \( g(x) \) will surpass \( f(x) \).
In summary, we can complete the sentences:
For small values of \( x \), the function with greater values is \( f(x) \). For instance, at \( x = 0 \), \( f(0) = 12 \) whereas \( g(0) = 3 \). As the value of \( x \) increases, however, \( g(x) \) eventually has greater values.