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