Question

look at the table of values for the functions f(x) = 2.5x + 2 and g(x) = 4^{x - 2}.
\begin{tabular}{|c|c|c|} hline x & f(x) & g(x) \\ hline 0 & 2 & $\frac{1}{16}$ \\ hline 1 & 4.5 & $\frac{1}{4}$ \\ hline 2 & 7 & 1 \\ hline 3 & 9.5 & 4 \\ hline 4 & 12 & 16 \\ hline end{tabular}
based on the values in the table, where does the equation f(x) = g(x) have a solution?
options: between x = 2 and x = 3, x = 3, between x = 3 and x = 4, x = 4

Explanation:

Step1: Analyze values at x=2

At \( x = 2 \), \( f(2)=7 \) and \( g(2)=1 \). So \( f(2)>g(2) \).

Step2: Analyze values at x=3

At \( x = 3 \), \( f(3)=9.5 \) and \( g(3)=4 \). So \( f(3)>g(3) \)? Wait, no, wait: Wait, \( g(3)=4^{3 - 2}=4^1 = 4 \), \( f(3)=2.5\times3 + 2=7.5 + 2 = 9.5 \). Wait, no, wait the table says \( f(3)=9.5 \), \( g(3)=4 \). Then at \( x = 3 \), \( f(3)=9.5 \), \( g(3)=4 \). At \( x = 4 \), \( f(4)=12 \), \( g(4)=16 \). Wait, so at \( x = 3 \), \( f(x)=9.5 \), \( g(x)=4 \) (so \( f(x)>g(x) \)), at \( x = 4 \), \( f(x)=12 \), \( g(x)=16 \) (so \( f(x)<g(x) \)). Wait, but earlier at \( x = 2 \), \( f(2)=7 \), \( g(2)=1 \) ( \( f>g \) ), \( x = 3 \): \( f=9.5 \), \( g=4 \) ( \( f>g \) ), \( x = 4 \): \( f=12 \), \( g=16 \) ( \( fg \), so no crossing there. Between \( x=3 \) and \( x=4 \): \( f \) goes from 9.5 to 12, \( g \) from 4 to 16. So \( f \) is below \( g \) at \( x=4 \), above at \( x=3 \), so crossing there.

Answer:

between \( x = 3 \) and \( x = 4 \)