Hilary T.

asked • 05/18/20# Find the rate of change

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?

## 1 Expert Answer

Mehek A. answered • 05/18/20

Engineer and Nanotech student making learning affordable

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

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Mehek A.

does t represent hours05/18/20