Question

Find the point where the following curve is steepest. (Round your answer to three decimal places.) y = 30/ (1 + 3e^(−3t)) for t ≥ 0

Answer 1

the graph is steepest when derivative has positive max ( or very negative min)

check where 2nd derivative is 0

Answer 2

\[\large {y = {30 \over 1 + 3e^{−3t}}}\]
\[\large {y'=-{30 \over \left( 1+ 3e^{-3t}\right)^2}\cdot-9e^{-3t}\\={270e^{-3t}\over \left( 1+ 3e^{-3t}\right)^2}}\]

\[\large {y''=-{810e^{-3t}\over \left( 1+ 3e^{-3t}\right)^2}+(-2)\cdot{270e^{-3t}\over \left( 1+ 3e^{-3t}\right)^3}\cdot-9e^{-3t}\\ = \color{blue}{810e^{-3t}\left( 3e^{-3t}-1 \right) \over\left( 1+ 3e^{-3t}\right)^3}}\]

Answer 3

when \(y''=0\): \[\large{3e^{-3t}-1=0\\ \implies e^{-3t}=1/3\\ \implies t=\color{green}{\ln \left( 3 \right)\over3}}\]

Answer 4

\[y=15\]the point is \((0.366,\ 15)\)