Question

50 БАЛЛОВ Найти наибольшее и наименьшее значения функции $y=\frac{x-3}{x^2+16}}$ на отрезке [-5; 5]

Answer 1

Ответ:

   $y_{min}=-\dfrac{1}{4}\ ,\ \ y_{max}=\dfrac{1}{16}\ ,\\\\y_{naimen.}=-\dfrac{8}{41}\ \ ,\ \ y_{naibol.}=\dfrac{2}{41}$  .

Решение.

$y=\dfrac{x-3}{x^2+16}\ \ ,\ \ x\in [-5\ ;\ 5\ ]\\\\\\y'=\dfrac{x^2+16-(x-3)\cdot 2x}{(x^2+16)^2}=\dfrac{-x^2+6x+16}{(x^2+16)^2}=0\ \ \Rightarrow \ \ \ -x^2+6x+16=0\\\\\\x^2-6x-16=0\ \ ,\ \ \ x_1=-2\ ,\ x_2=8\ \ \ (teorema\ Vieta)\\\\\\\dfrac{-(x+2)(x-8)}{(x^2+16)^2}=0\\\\\\znaki\ y'(x):\ \ \ ---(-2)+++(8)---\\{}\qquad \qquad \qquad \quad \ \searrow \, \ \ (-2)\ \ \nearrow \ \ \ (8)\ \ \searrow \\{}\qquad \qquad \qquad \qquad \qquad min\ \ \ \qquad max$

$x_{min}=-2\ \ ,\ \ y_{min}=y(-2)=\dfrac{-2-3}{4+16}=-\dfrac{5}{20}=-\dfrac{1}{4}\ \ ,\ \ -2\in [-5\ ;\ 5\ ]\\\\\\x_{max}=8\ \ ,\ \ \ y_{max}=y(8)=\dfrac{8-3}{64+16}=\dfrac{5}{80}=\dfrac{1}{16}\ ,\ \ 8\notin [-5\ ;\ 5\ ]\\\\y(-5)=\dfrac{-8}{41}\ \ ,\ \ y(5)=\dfrac{2}{41}\\\\\\-\dfrac{8}{41}<-\dfrac{1}{4}<\dfrac{2}{41}\ \ \ \Rightarrow \ \ \ y_{naimen.}=y(-5)=-\dfrac{8}{41}\ \ ,\ \ y_{naibol.}=y(5)=\dfrac{2}{41}$