What are the critical points of s(t)=(e^t-2)^4/(e^t+7)^5s(t)=(et−2)4(et+7)5?

1 Answer
Harish Chandra Rajpoot
Jul 2, 2018

t=\ln2, \ln38t=ln2,ln38

Explanation:

Given function

s(t)=\frac{(e^t-2)^4}{(e^t+7)^5}s(t)=(et−2)4(et+7)5

\frac{d}{dt}(s(t))=\frac{(e^t+7)^5\frac{d}{dt}(e^t-2)^4-(e^t-2)^4\frac{d}{dt}(e^t+7)^5}{((e^t+7)^5)^2}ddt(s(t))=(et+7)5ddt(et−2)4−(et−2)4ddt(et+7)5((et+7)5)2

=\frac{(e^t+7)^{5}4(e^t-2)^3e^t-(e^t-2)^{4}5(e^t+7)^4e^t}{(e^t+7)^10}=(et+7)54(et−2)3et−(et−2)45(et+7)4et(et+7)10

=\frac{e^t(e^t+7)^{4}(e^t-2)^3(4(e^t+7)-5(e^t-2))}{(e^t+7)^10}=et(et+7)4(et−2)3(4(et+7)−5(et−2))(et+7)10

=\frac{e^t(e^t-2)^3(38-e^t)}{(e^t+7)^6}=et(et−2)3(38−et)(et+7)6

For critical points, we have

\frac{d}{dt}(s(t))=0ddt(s(t))=0

\frac{e^t(e^t-2)^3(38-e^t)}{(e^t+7)^6}=0et(et−2)3(38−et)(et+7)6=0

(e^t-2)^3(38-e^t)=0\quad (\because \ \ e^t>0)

e^t-2=0\ or \ 38-e^t=0

e^t=2\ or \ e^t=38

t=\ln2\ or \ t=\ln38

t=\ln2, \ln38

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