Answer to Question #106008 in Calculus for Parul

Question

Trace the curve y= cube root of x^2 -1 and state all the properties you use to trace it.

Answer 1

$y=\sqrt[3]{x^2-1}$

1) The domain $x\in(-\infty;\infty)$ .

 2) Evenness, oddity

$y(-x)=\sqrt[3]{(-x)^2-1}=\sqrt[3]{x^2-1}=y(x)$

graph is symmetric about the axis $Oy$.

3)  Continuity

the function is continuous $x\in (-\infty;\infty)$

 4) Periodicity

the function is not periodic

 5) Intersection points with coordinate axes

$Ox: y=0\\ \sqrt[3]{x^2-1}=0\\ x^2-1=0\\ x^2=1\\ x_1=1, x_2=-1\\ A(1,0), B(-1,0)\\ Oy:x=0\\ y=\sqrt[3]{0^2-1}=\sqrt[3]{-1}=-1\\ C(0,-1)$

6) Let's check the extremum

$y'=((x^2-1)^\frac{1}{3})'=\frac{2}{3}\cdot x\cdot(x^2-1)^{-\frac{2}{3}}\\ y'=0\\ \frac{2}{3}\cdot x\cdot(x^2-1)^{-\frac{2}{3}}=0\\ x_1=0, x_2=1, x_3=-1\\ x\in(-\infty;-1)\cup(-1;0], y\searrow\\ x\in [0;1)\cup(1,\infty),y\nearrow\\ x=0-min\\ y_{min}=-1$

7) Convex intervals

$y''=\frac{2}{3}(x^2-1)^{-\frac{2}{3}}-\frac{8}{9}\cdot x^2\cdot(x^2-1)^{-\frac{5}{3}}\\ y''=0\\ \frac{2}{3}(x^2-1)^{-\frac{2}{3}}-\frac{8}{9}\cdot x^2\cdot(x^2-1)^{-\frac{5}{3}}=0\\ (x^2-1)^{-\frac{5}{3}}(x^2-1-\frac{4}{3}x^2)=0\\ x_1=-1,x_2=1, \frac{1}{3}x^2=-1\\ x\in (-\infty;-1)\cup(1,\infty)$

since the second derivative is at this interval

$y''<0$ then

$y$ is convex upward $\cap$

$x\in (-1,1)$

since the second derivative is at this interval

$y''>0$ then

$y$ is convex down $\cup$

$x=1, x=-1$ are inflection points

8) Asymptotes

 there are no vertical asymptotes

 find oblique, horizontal asymptote $y=kx+b$

$k=\lim\limits_{ x\to\infty}\frac{f(x)}{x}=\lim\limits_{ x\to\infty}\frac{\sqrt[3]{x^2-1}}{x}=0\\ b=\lim\limits_{ x\to\infty}(f(x)-kx)=\lim\limits_{ x\to\infty}(\sqrt[3]{x^2-1}-0)=\infty$

there are no oblique, horizontaltical asymptote $y=kx+b$

9)

$\lim\limits_{x\to -\infty}\sqrt[3]{x^2-1}=+\infty\\ \lim\limits_{x\to\infty}\sqrt[3]{x^2-1}=+\infty$