Question

which function could produce the graph shown below? a. $f(x) = 4e^{-0.5x}$ b. $f(x) = 4e^{-0.05x}$ c. $f(x) = 4e^{0.5x}$ d. $f(x) = 4e^{0.05x}$

Explanation:

Step1: Analyze the graph's trend

The graph is a decreasing exponential curve (since as \( x \) increases, \( y \) decreases). For exponential functions of the form \( f(x) = ae^{kx} \), if \( k < 0 \), the function is decreasing; if \( k > 0 \), it's increasing. So we can eliminate options C (\( k = 0.5>0 \)) and D (\( k = 0.05>0 \)) as they are increasing functions.

Step2: Analyze the rate of decrease

Now we have options A (\( k=-0.5 \)) and B (\( k = - 0.05 \)). The magnitude of \( |k| \) determines the rate of decrease: a larger \( |k| \) means a faster decrease. Looking at the graph, the curve decreases relatively slowly (since it takes a long time to go from \( y = 4 \) to near \( y = 0 \) as \( x \) goes from 0 to 35). The function \( f(x)=4e^{-0.5x} \) (option A) would decrease very rapidly (e.g., at \( x = 2 \), \( f(2)=4e^{-1}\approx1.47 \); at \( x = 4 \), \( f(4)=4e^{-2}\approx0.54 \)), which is too fast. The function \( f(x)=4e^{-0.05x} \) (option B) decreases more slowly: at \( x = 10 \), \( f(10)=4e^{-0.5}\approx4\times0.6065 = 2.426 \); at \( x = 20 \), \( f(20)=4e^{-1}\approx1.47 \); at \( x = 30 \), \( f(30)=4e^{-1.5}\approx4\times0.2231 = 0.892 \), which matches the slow - decreasing trend of the graph.

Answer:

B. \( f(x)=4e^{-0.05x} \)