найдите точку минимума функции [tex]- \frac{x}{ x^{2} +81} [/tex]. Created by bpan93rus. algebra-ru

tex]...

baraybek Производная(-x/(x2+81))=-(x2+81-2x2)/(x2+81)2=(x2-81)/(x2+81)2
(x2-81)/(x2+81)2=0
x2=81
x=+-9

o1l7eg17 $y=- \frac{x}{x^2+81}$

$y'=(- \frac{x}{x^2+81})'=-( \frac{x}{x^2+81})'= \frac{x'(x^2+81)-x(x^2+81)'}{(x^2+81)^2}=$

$=- \frac{-x(x^2+81)'+(x^2+81)}{(x^2+81)^2}=-\frac{81+x^2-2x^2}{(x^2+81)^2}=-\frac{81-x^2}{(x^2+81)^2}$

$- \frac{81-x^2}{81+x^2}=0$

$\frac{x^2-81}{(x+81)^2}=0$

$x^2-81=0$
$(x-9)(x+9)=0$
$x-9=0$ $x+9=0$
$x_1=9$ $x_2=-9$

Интервалы монотонности смотри в вложении.

Ответ: т. 9 - точка минимума.

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