Find the rate of change

Question

given c(t) = e^(-0.06t)-e^(-0.44t), find the rate of change after 2h.

At what time does the rate begin to decrease?

Answer 1

Assuming t represents hours

c(t) = e^-0.06t-e^-0.44t

rate of change means the derivative which can be denoted as c'(t)

c'(t) = -0.06e^-0.06t+0.44e^-0.44t

Rate of Change after two hours:

c'(2) = 0.23571 units/hour

When we solve for the roots of c'(t) or set c'(t) = 0, we can figure out where there is a sign change.

When a derivative is negative, that means the original function c(t) is decreasing.

Setting c'(t) = -0.06e^-0.06t+0.44e^-0.44t = 0

We figure out that t = (ln(0.44)−ln(0.06)) / 0.38 = 5.24323 hours

After doing the first derivative time test we see that our y-values for our derivative is less than zero after t = 5.24323 hours.

your y-values for c'(t) become more negative after 5.24323 hours , which means this is where c(t) is decreasing. t > 5.24323 is where the function is decreasing