## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises: 77

#### Answer

28.57 , 85.71 The graph for the drug response curve is as shown below:

#### Work Step by Step

Given: $S(t)=At^{p}e^{-kt}$ Take first and second derivative of the function $S'(t)=Apt^{p-1}e^{-kt}-kApt^{p}e^{-kt}$ $S'(t)=(p-kt)t^{p-1}e^{-kt}A=At^{p}e^{-kt}(\frac{p}{t}-k)$ and $S''(t)=At^{p}e^{-kt}[(\frac{p}{t}-k)^{2}-\frac{p}{t^{2}}]$ Substitute the various values such as: $A = 0.01, p = 4, k = 0.07$ we get a graph as shown below: We observe from the graph that the rate of increase of drug level in blood stream is greatest after 28.57 minutes and rate of decrease is greatest at 85.71 min. Hence, the inflection points are: 28.57 , 85.71 .

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