Question

Исследуйте на монотонность, экстремумы и постройте график функции. Помогите срочно

Answer 1

$y=\dfrac{2x}{1-x^2}\\\\1)\ \ ODZ:\ \ x\ne \pm 1\\\\2)\ \ OY:\ x=0\ \ \to \ \ y(0)=\dfrac{2\cdot 0}{1}=0\ \ ,\ \ O(0;0)\\\\OX:\ \ y=0\ \ \to \ \ \ \dfrac{2x}{1-x^2}=0\ \ ,\ \ x=0\ \ ,\ \ O(0;0)\\\\3)\ \ y<0\ ,\ esli\ \ \dfrac{2x}{1-x^2}<0\ \ ,\ \ \dfrac{2x}{(1-x)(1+x)}<0\ \ ,\\\\\\znaki\ y(x):\ \ +++(-1)---(0)+++(1)---\\\\y<0\ ,\ esli\ \ x\in (-1;\, 0\ )\cup (\ 1\ ;+\infty \, )\\\\y>0\ ,\ esli\ \ x\in (-\infty ;-1\ )\cup (\ 0\ ;\ 1\ )$

$4)\ \ Asimptotu.\\\\\lim\limits_{x \to -1}\dfrac{2x}{1-x^2}=[\frac{-2}{0}]=\infty \ \ ,\ \ \lim\limits_{x \to 1}\dfrac{2x}{1-x^2}=[\frac{2}{0}]=\infty \ \ \ \Rightarrow \\\\\underline {y=-1\ \ \ i\ \ \ \ y=1}\ \ \ \ \ vertikalnue$

$y=kx+b\ \ \ naklonnaya\ asimhtjta\\\\k=\lim\limits_{x \to \infty }\dfrac{y(x)}{x}=\lim\limits_{x \to \infty}\dfrac{2x}{x(1-x^2)}=\lim\limits_{x \to \infty}\dfrac{2x}{-x^3}=0\\\\b=\lim\limits_{x \to \infty }(y-kx)=\lim\limits_{x \to \infty }\dfrac{2x}{1-x^2}=0\ \ \ \to \ \ \ \underline {y=0\ }\ \ \ gorizontalnaya\ asimptota$

$5)\ \ y'=\dfrac{2(1-x^2)+2x\cdot 2x}{(1-x^2)^2}=\dfrac{2(1+x^2)}{(1-x^2)^2}\ne 0\\\\y'(x)>0\ \ dlya\ \ x\ne \pm 1\ \ \Rightarrow \\\\y(x)\ vozrastaet\ \ pri\ \ x\in ODZ\\\\\\6)\ \ y''=\dfrac{4x(1-x^2)^2+2(1+x^2)\cdot 2\cdot (1-x^2)\cdot 2x}{(1-x^2)^4}=\\\\\\=\dfrac{4x(1-x^2)\cdot (1-x^2+2(1+x^2))}{(1-x^2)^4}=\dfrac{4x(3+x^2)}{(1-x^2)^3}=0\ \ \to \ \ \ x=0\ ,\ x\ne \pm 1\\\\O(\ 0\ ;\ 0\ )\ \ -\ \ tochka\ peregiba$