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

1 Answer

Answer:

#t=\ln2, \ln38#

Explanation:

Given function

#s(t)=\frac{(e^t-2)^4}{(e^t+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}#

#=\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}#

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

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

For critical points, we have

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

#\frac{e^t(e^t-2)^3(38-e^t)}{(e^t+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#