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

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Answer 1 · 2018-07-02T20:02:40

$t=\ln2, \ln38$

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

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

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

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

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

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

$t=\ln2\ or \ t=\ln38$

$t=\ln2, \ln38$

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