Answer:
exponential / half-life function:
[tex]f(x) = 384\!\left(\dfrac{1}{2}\right)^{\!\!x-1}[/tex]
Step-by-step explanation:
We are given a table of function inputs and their corresponding outputs:
[tex]\begin{array}{ c | c } x & f(x) \\\cline{1-2} 1 & 384 \\\cline{1-2} 2 & 192 \\\cline{1-2} 3 & 96 \\\cline{1-2} 4 & 48 \\\cline{1-2} 5 & 24 \end{cases}[/tex]
To analyze what kind of function this is, we can look at the slope between the given points.
Between x = 1 and x = 2:
[tex]m_{1\to 2} = \dfrac{f(2) - f(1)}{2-1} = f(2)-f(1)[/tex]
[tex]m_{1\to 2} = 384 - 192[/tex]
[tex]m_{1\to 2} = 192[/tex]
Between x = 2 and x = 3:
[tex]m_{2\to 3} = \dfrac{f(3) - f(2)}{3-2} = f(3)-f(2)[/tex]
[tex]m_{2\to 3} = 192-96[/tex]
[tex]m_{2\to 3} = 96[/tex]
Between x = 3 and x = 4:
[tex]m_{3\to 4} = \dfrac{f(4) - f(3)}{4-3} = f(4)-f(3)[/tex]
[tex]m_{3\to 4} = 96-48[/tex]
[tex]m_{3\to 4} = 48[/tex]
We can see that each slope is half the slope over the previous Δx = 1:
[tex](m_{1\to 2} = 192) \ \stackrel{\longrightarrow}{_{\times 1/2}} \ (m_{2\to 3} = 96)[/tex]
[tex](m_{2\to 4} = 96) \ \stackrel{\longrightarrow}{_{\times 1/2}} \ (m_{3\to 4} = 48)[/tex]
This means that the function changes relative to its current value, which is the definition of an exponential function.
In this case, since the function halves every Δx = 1, we can create the half-life function with initial value (1, 384):
[tex]\boxed{f(x) = 384\!\left(\dfrac{1}{2}\right)^{\!\!x-1}}[/tex]
Notice that the exponent of 1/2 is (x - 1) because the initial point is centered at x = 1 (so we have to shift the graph to the right 1).