Question

A population of bacteria can be modeled by the function f(t)= 400(0.98)t, where t is time in hours. What is the rate of change in the function?

Answer 1

Answer:

$\dfrac{d(f(t))}{dt}=-8.08(0.98)^t$        

Step-by-step explanation:

We are given the following function:

$f(t)= 400(0.98)^t$

The function gives the population of bacteria in time, t where t is in hours.

We have to find the rate of change in function.

Rate of change =

$\dfrac{d(f(t))}{dt} = \dfrac{d}{dt}(400(0.98)^t)\\\\\dfrac{d(f(t))}{dt} = 400(\ln 0.98)(0.98)^t\\\\\dfrac{d(f(t))}{dt}=-8.08(0.98)^t$

is the required rate of change in function.

Differentiation property used:

$\dfrac{d}{dx}(a^x) = (\ln a)a^x$

Answer 2

Answer:

C). decrease 2% per hour