Answer to Question #104989 in Calculus for Hitesh

Question

Trace the curve y =
\sqrt[3]{ {x}^{2} - 1}
3

x
2
−1
​

and state all the properties you use to trace it

Answer 1

Solution:

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

1.Thej domain :$~x\in(-\infty,+\infty)$

2. Intersection points with coordinate axes:

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

the graph is symmetric about the axis Oy

4. The function is continuous

5. We find the critical points: $y'=\frac{1}{3}(x^2-1)^{-\frac{2}{3}}\cdot2x\to x=0, x=1,x=-1$

$x\in(-\infty,-1)\cup(-1,0), y' \lt 0\to y\searrow\\ x\in(0,1)\cup(1,\infty), y' \gt 0 \to y\nearrow\\$

Then $x=0$  is the minimum point $y=-1,x=0$

6. Find the inflection points:

$y''=\frac{2}{3}(x^2-1)^{-\frac{2}{3}}+\frac{2}{3}x\cdot(-\frac{2}{3})(x^2-1)^{-\frac{5}{3}} =$

$\frac{2}{3}(x^2-1)^{-\frac{5}{3}}(x^2-1-\frac{4}{3}x^2) =\frac{2}{3}(x^2-1)^{-\frac{5}{3}}(-\frac{1}{3}x^2-1)\\~\\ x\in(-\infty,-1)\cup(1,+\infty)\to y'' \lt 0\\ x\in (-1,1)\to y'' \gt 0$

7. There are no vertical asymptotes horizontal asymptote:$y=kx+b~~~\to~~~ k=\lim_{x\to\infty} \frac{\sqrt[3]{x^2-1}}{x}=0\\ b=\lim_{x\to\infty} (\sqrt[3]{x^2-1}-0\cdot x)=\infty$

there are no horizontaltical asymptotes

8. Intervals for the range:

$y \gt 0 \to x^2-1 \gt 0,~ x^2 \gt 1 \to x\in(-\infty;-1)\cap (1;\infty)\\ y \lt 0\to x\in(-1;1)$

9. Behavior at infinity: $\begin{bmatrix} x\to-\infty, y\to+\infty\\ x\to+\infty, y\to+\infty \end{bmatrix}$