Question

Trace the curve y^3 = x^2 −1 , and state all the properties you use to trace it.

Answer 1

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

1 thej domain $x\in(-\infty;+\infty)$

2 intersection points with coordinate axes

$Ox: y=0\\ x^2-1=0\\ x_1=1, x_2=-1\\ A(1;0), B(-1;0)\\ Oy:x=0\\ y^3=-1\\ y=-1\\ C(0;-1)$

3 symmetry

$y^3=(-x)^2-1=x^2-1$

the graph is symmetric about the axis $Oy$

4 the function is continuous

5 we find the critical points

$y'=\frac{1}{3}(x^2-1)^{-\frac{2}{3}}\cdot2x\\ x=0, x=1,x=-1\\ x\in(-\infty,-1)\cup(-1,0), y'<0, y\searrow\\ x\in(0,1)\cup(1,\infty), y'>0,y\nearrow\\$

Then $x=0$ is the maximum point

$y=-1,x=0$

6 find the inflection points

$y''=\frac{2}{3}(x^2-1)^{-\frac{2}{3}}+\frac{2}{3}x\cdot(-\frac{2}{3})(x^2-1)^{-\frac{5}{3}}=\\ =\frac{2}{3}(x^2-1)^{-\frac{5}{3}}(x^2-1-\frac{4}{3}x^2)=\\ =\frac{2}{3}(x^2-1)^{-\frac{5}{3}}(-\frac{1}{3}x^2-1)\\ x\in(-\infty,-1)\cup(1,+\infty), y''<0\\ x\in (-1,1), y''>0$

7 there are no vertical asymptotes

 horizontal asymptote

$y=kx+b\\ k=\lim_{x\to\infty} \frac{\sqrt[3]{x^2-1}}{x}=0\\ b=\lim_{x\to\infty} (\sqrt[3]{x^2-1}-0\cdot x)=\infty$

there are no horizontaltical asymptotes

8 intervals for the range

$y>0, x^2-1>0\\ x^2>1\\ x\in(-\infty;-1)\cap (1;\infty)\\ y<0, x\in(-1;1)$

9 behavior at infinity

$x\to-\infty, y\to+\infty\\ x\to+\infty, y\to+\infty$