Question

Найдите наименьшее значение функции y=(x−11)e^x−10 на отрезке [9;11].

Answer 1

$\displaystyle f(x)=(x-11)e^x-10=xe^x-11e^x-10;$
$\displaystyle f'(x)=\frac{df(x)}{dx}=\frac{d}{dx}\left(xe^x-11e^x-10\right)=\frac{d}{dx}\left(xe^x\right)-\frac{d}{dx}\left(11e^x\right)-\frac{d}{dx}\left(10\right)=\frac{xd\left(e^x\right)+e^xdx}{dx}-11\frac{d}{dx}\left(e^x\right)-0=\frac{xe^xdx}{dx}+\frac{e^xdx}{dx}-11e^x=xe^x+e^x-11e^x=xe^x-10e^x=e^x(x-10);$

$\displaystyle 0=f'(x)=e^x(x-10);$
$\displaystyle \forall x\in\mathbb{R}: e^x\neq 0 \implies \Big(e^x\left(x-10\right)=0 \iff x-10=0\Big);$
$\displaystyle x-10=0\implies x=10;$

$\displaystyle f(9)=(9-11)e^9-10=-2e^9-10;$
$\displaystyle f(10)=(10-11)e^{10}-10=-e^{10}-10;$
$\displaystyle f(11)=(11-11)e^{11}-10=-10;$

$\displaystyle f(9)\stackrel{?}{=}f(10);$
$\displaystyle -2e^9-10\stackrel{?}{=}-e^{10}-10;$
$\displaystyle -2e^9\stackrel{?}{=}-e^{10};$
$\displaystyle 2e^9\stackrel{?}{=}e^{10};$
$\displaystyle ln(2e^9)\stackrel{?}{=}ln(e^{10});$
$\displaystyle ln(2)+ln(e^9)\stackrel{?}{=}10;$
$\displaystyle ln(2)+9\stackrel{?}{=}10;$
$\displaystyle ln(2)\stackrel{?}{=}1;$
$\displaystyle e^{ln(2)}\stackrel{?}{=}e^1;$
$\displaystyle 2\stackrel{?}{=}e;$
$\displaystyle 2<e\approx 2.72;$
$\displaystyle 2<e \implies -2>-e\implies -2e^9>-e^{10}\implies f(9)>f(10);$

$\displaystyle f(10)\stackrel{?}{=}f(11);$
$\displaystyle -e^{10}-10\stackrel{?}{=}-10;$
$\displaystyle -e^{10}-10+10\stackrel{?}{=}0;$
$\displaystyle -e^{10}<0 \implies f(10)<f(11);$

$\displaystyle f(9)>f(10)\land f(10)<f(11) \implies \min_{x\in[9;11]}f(x)=f(10)=\boxed{-e^{10}-10}\phantom{.}.$