Question

интервалы возрастания и убывания

Answer 1

Ответ:

$y^3=(x+1)^2(x-1)\ \ \ \Rightarrow \ \ \ y=\sqrt[3]{(x+1)^2(x-1)}\\\\y'=\dfrac{1}{3}\cdot \Big((x+1)^2(x-1)\Big)^{-\frac{2}{3}}\cdot \Big(2(x+1)(x-1)+(x+1)^2\Big)=\\\\=\dfrac{2(x^2-1)+(x+1)^2}{3\, \sqrt[3]{\Big(x+1)^2(x-1)\Big)^2}}=\dfrac{2x^2-2+x^2+2x+1}{3\sqrt[3]{(x+1)^4(x-1)^2}}=\dfrac{3x^2+2x-1}{3\sqrt[3]{(x+1)^4(x-1)^2}}=0\\\\\\3x^2+2x-1=0\ \ ,\ \ x\ne \pm 1\\\\D=16\ \ ,\ \ x_1=-1\notin ODZ\ ,\ \ x_2=\dfrac{1}{3}$

$\dfrac{3(x+1)(x-\frac{1}{3})}{3\sqrt[3]{(x+1)^4(x-1)^2}}=0\\\\\\znaki\ y'(x):\ \ \ +++(-1)---(\frac{1}{3})+++(1)+++\\{}\qquad \qquad \qquad \quad \nearrow \ \ \ (-1)\ \ \ \searrow \ \ \, (\frac{1}{3})\ \ \ \nearrow \ \, (1)\ \ \nearrow \\{}\qquad \qquad \qquad \qquad \quad \ max\qquad \quad \ min$

$y(x)\ vozrastaet\ \ pri\ \ x\in (-\infty ;-1\ ]\cup [\ \frac{1}{3}\ ;+\infty \, )\\\\y(x)\ \ ybuvaet\ \ pri\ \ x\in [-1\ ;\ \frac{1}{3}\ ]\\\\x_{max}=-1\ \ ,\ \ \ y_{max}=y(-1)=0\\\\x_{min}=\dfrac{1}{3}\ \ ,\ \ \ y_{min}=y\Big(\dfrac{1}{3}\Big)=-\sqrt[3]{\dfrac{32}{27}}\approx -1,06$