Построить график функции очень подробно и с разъяснением y=2/(x^2+4). Created by Igor171717. algebra-ru

(x^2+4)...

sedinalana $y= \frac{2}{x^2+4} \\\\1)\; \; ODZ:\; \; x\in (-\infty ,+\infty )\\\\2)\; \; y(-x)= \frac{2}{(-x)^2+4}=\frac{2}{x^2+4} \quad \to \; \; chetnaya\\\\3)\; \; OX:\; \; y=0\; \; \to \; \; \frac{2}{x^2+4} =0\; \; \Rightarrow \; \; x\in \varnothing \\\\(net\; peresecheniya\; s\; OX)\\\\3)\; \; OY:\; \; x=0\; \; \to \; \; y(0)= \frac{2}{4}=\frac{1}{2}\\\\4)\; \; Kriticheskie\; tochki:\\\\y'=\frac{-2\cdot 2x}{(x^2+4)^2}=-\frac{4x}{(x^2+4)^2}=0\; \; \to \; \; x=0\; ,\; y(0)=\frac{1}{2}\\\\5)\; \; Znaki\; y'(x):\;$

$++++++(0)------\\\\.\qquad \; \; \; \nearrow \; \quad (0)\; \; \; \; \; \searrow \\\\vozrastaet:x\in (-\infty ,0)\; ,\; \; ybuvaet:\; \; x\in (0,+\infty )\\\\x_{max}=0\; ,\; y_{max}=\frac{1}{2}\\\\6)\; \; y''= \frac{-4(x^2+4)+4x\cdot 2\cdot (x^2+4)\cdot 2x}{(x^2+4)^2} = \frac{4(3x^2-4)}{(x^2+4)^3} =0\\\\3x^2-4=0\; ,\; \; x^2=\frac{4}{3}\; \; \to \; \; x=\pm \frac{2}{\sqrt3}\approx \pm 1,15\\\\7)\; \; Znaki\; \; y''(x):$

$+++(-\frac{2}{\sqrt3})---(\frac{2}{\sqrt3})+++\\\\Vognytaya:\; \; x\in (-\infty ,\frac{2}{\sqrt3})\cup (\frac{2}{\sqrt3},+\infty )\\\\Vupyklaya:\; \; x\in (-\frac{2}{\sqrt3},\frac{2}{\sqrt3}) \\\\ y(\pm \frac{2}{\sqrt3})= \frac{2}{ \frac{4}{3}+4}=\frac{3}{8}=0,375 \\\\7)\; \; y(x)=\frac {2}{x^2+4}\ \textgreater \ 0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\y(x)\in (0,+\infty )\\\\8)\; \; Asimptotu:\; \; \lim\limits _{x \to \infty}y(x)= \lim\limits _{x \to \infty}\frac{2}{x^2+4}=[\frac{2}{\infty }]=0\; \; \to \\\\asimptota\; \; y=0
$

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Verified answer

Y=2/(x²+4)
D(y)∈R
y(-x)=2/((-x)²+4)=2/(x²+4) четная
y`=-4x/(x²+4)²=0
x=0
+ _
----------------(0)--------------------
возр max убыв
y(0)=1/2
y``(x)=(-4(x²+4)²-2(x²+4)*2x*(-4x))/(x²+4)^4=-4(x²+4)(x²+4-4x²)/(x²+4)^4=
=-4(x²+4)(4-3x²)/(x²+4)^4=0
4-3x²=0
x=-2/√3 x=2/√3
_ + _
-----------------(-2/√3)--------------------(2/√3)-----------------
выпук вверх вогн вниз выпук вверх
y(-2/√3)=2/(4/3+4)=3/8
y(2/√3)=3/8

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