Question 90·Medium·Math

Compute $\dfrac{f(1)}{f(0)}$, $\dfrac{f(2)}{f(1)}$, and $\dfrac{f(3)}{f(2)}$. What number do you keep multiplying by as $x$ increases by 1?

Once you know the starting value $f(0)$ and the factor you multiply by each time $x$ increases by 1, think of a function of the form $a\cdot b^x$. What are $a$ and $b$ here?

After you write your function using $\left(\tfrac{1}{2}\right)^x$, rewrite it using a power of 2 so you can compare it directly with the answer choices.

In Desmos, type the four options as separate functions: y = 5*2^(x+1), y = 5*2^x, y = 5*2^(-(x+1)), and y = 5*2^(-x).

For each function, click the gear icon next to it and select "Table". In each table, enter the x-values 0, 1, 2, and 3.

Compare the y-values in each Desmos table with the problem’s values $5, \dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}$ at $x=0,1,2,3$. The function whose y-values exactly match all four values is the correct choice.

Write the $f(x)$ values in order: $5,\; \dfrac{5}{2},\; \dfrac{5}{4},\; \dfrac{5}{8}$.

Now compare consecutive terms:

• From $5$ to $\dfrac{5}{2}$: you multiply by $\dfrac{1}{2}$.
• From $\dfrac{5}{2}$ to $\dfrac{5}{4}$: you multiply by $\dfrac{1}{2}$ again.
• From $\dfrac{5}{4}$ to $\dfrac{5}{8}$: you again multiply by $\dfrac{1}{2}$.

So every time $x$ increases by 1, $f(x)$ is multiplied by $\dfrac{1}{2}$. This is a constant ratio, which means an exponential function of the form $a\cdot b^x$ with $b=\dfrac{1}{2}$. The starting value (when $x=0$) is $5$, so $a=5$.

Using the initial value and the common ratio, we can write

• Initial value (when $x=0$): $f(0)=5$.
• Each step in $x$ multiplies the output by $\dfrac{1}{2}$.

So the function can be written as

This matches the pattern in the table (you can quickly check: $x=0$ gives $5$, $x=1$ gives $5\cdot \tfrac{1}{2}=\tfrac{5}{2}$, etc.).

The answer choices all use expressions of the form $2^{\text{(something)}}$, so rewrite $\left(\tfrac{1}{2}\right)^x$ using a power of 2:

• Note that $\dfrac{1}{2} = 2^{-1}$.
• Therefore,

Substitute this into the equation from the previous step:

This matches choice D, so the correct equation is $f(x)=5(2^{-x})$.