Question

For $0\le t<\infty ,$ the maximum value of the function $f\left(t\right)=e^{-t}-2e^{-2t}$ occurs at

• A. $t=lo{g}_{e}4$
• B. $t=lo{g}_{e}2$
• C. $t=0$
• D. $t=lo{g}_{e}8$

Answer 1

The correct option is A $t=lo{g}_{e}4$

$f\left(t\right)=e^{-t}-2e^{-2t},0\le t<\infty$
$f^{\prime }\left(t\right)=-e^{-t}+4e^{-2t}$
For critical points $f^{\prime }\left(t\right)=0\Rightarrow$ Either
$e^{-t}(4e^{-t}-1)=0$
$e^{-t}=0$
$t=\infty$ (Not possible)
or $4e^{-t}-1=0$
$e^{-t}=\frac{1}{4}$
$e^{-t}=4$
$t=lo{g}_{e}4$
$f^{\prime \prime }\left(t\right)=e^{-t}-8e^{-2t}$
$f^{\prime \prime }\left(lo{g}_{e}4\right)=e^{-lo{g}_{e}4}-8e^{-2lo{g}_{e}4}$
$=\frac{1}{4}-8\left(\frac{1}{16}\right)=-\frac{1}{4}<0$
$f\left(t\right)$ has maximum value at
$t=lo{g}_{e}4$