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

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

$displaystyle 0=f(x)=e^x(x-10);$
$displaystyle forall xinmathbb{R}: e^xneq 0 implies Big(e^xleft(x-10right)=0 iff x-10=0Big);$
$displaystyle x-10=0implies x=10;$

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

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

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

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