Найдите точки экстремума функции: y=(2x+1)*e^-x. Created by khrushkovam. algebra-ru

Найдите точки экстремума функции: y=(2x+1)*e^-x...

khrushkovam @khrushkovam

July 2022 1 7 Report

Найдите точки экстремума функции: y=(2x+1)*e^-x

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$y=(2x+1)\cdot e^{-x}\\\\y'=2e^{-x}-(2x+1)e^{-x}=e^{-x}\cdot (2-2x-1)=e^{-x}\cdot (1-2x)=0\\\\a)\; \; e^{-x}\ne 0\; ,\; t,k,\; \; e^{-x}\ \textgreater \ 0\\\\b)\; \; 1-2x=0\; ,\; \; x=\frac{1}{2}\\\\Znaki\; y':\; \; \; ---(\frac{1}{2})+++\\\\x_{min}=\frac{1}{2}\\\\y_{min}=2\cdot e^{-\frac{1}{2}}=\frac{2}{\sqrt{e}}\approx 1,21$

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