Answer to Question #104509 in Calculus for Aman

Question

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

Answer 1

The curve $y^3=x^2-1$ coincides with the graph of the function $\quad \quad y(x)={ \left( { x }^{ 2 }-1 \right) }^{ \frac { 1 }{ 3 } }\quad \quad$

1) The function is defined on $\left( -\infty ,+\infty \right)$ .

2) The function is continuous in the domain of definition, since it is elementary.

3) The function is even , so the graph of the function is symmetric about the axis 0Y.

4) ${ y }^{ ' }(x)=\frac { 2 }{ 3 } x{ \left( { x }^{ 2 }-1 \right) }^{ -\frac { 2 }{ 3 } },x\in \left( -\infty ,-1 \right) \cup \left( -1,1 \right) \cup \left( 1,+\infty \right)$ .

5)By definition , the line $x=a$ is the vertical tanget to the graph of the function at the point $(a,y(a))$ if the function is continuous and $\lim _{ x\rightarrow a }{ \frac { y(x)-y(a) }{ x-a } } =\infty$

$y(1)=0, y(-1)=0$ $\lim _{ x\rightarrow 1 }{ \frac { { \left( { x }^{ 2 }-1 \right) }^{ \frac { 1 }{ 3 } }-0 }{ x-1 } =\lim _{ x\rightarrow 1 }{ \frac { { \left( x+1 \right) }^{ \frac { 1 }{ 3 } } }{ { \left( x-1 \right) }^{ \frac { 2 }{ 3 } } } } =\infty } ,$ $\lim _{ x\rightarrow -1 }{ \frac { { \left( { x }^{ 2 }-1 \right) }^{ \frac { 1 }{ 3 } }-0 }{ x+1 } =\lim _{ x\rightarrow -1 }{ \frac { { \left( x-1 \right) }^{ \frac { 1 }{ 3 } } }{ { \left( x+1 \right) }^{ \frac { 2 }{ 3 } } } } =\infty } .$ Thus, the function $y(x)$ is not differentiable at the points $x=-1, x=1$. The lines $x=-1,x=1$ are the vertical tangents to the function graph.

6) The function increases in the intervals $\ \left( 0,1 \right) ,\ \left( 1,+\infty \right)$ , since $y'(x)>0$

The function decreases in the intervals $\left( -\infty ,-1 \right) , \left( -1,0 \right)$, since $y'(x)<0$

7) From 6) it follows that x=0 is a point of local and global minimum $\ { y }_{ min }=y(0)=-1$ . $\lim _{ x\rightarrow +\infty }{ { \left( { x }^{ 2 }-1 \right) }^{ \frac { 1 }{ 3 } }=+\infty }$ The range of function is $\left[- 1,+\infty \right)$

8) "\\quad { y }^{ " }(x)=\\frac { 2 }{ 3 } { \\left[ { \\ \\left( { x }^{ 2 }-1 \\right) }^{ -\\frac { 2 }{ 3 } }-\\frac { 4 }{ 3 } { x }^{ 2 }{ \\left( { x }^{ 2 }-1 \\right) }^{ -\\frac { 5 }{ 3 } } \\right] }" $=-\frac { 2 }{ 9 } { \left( { x }^{ 2 }-1 \right) }^{ -\frac { 5 }{ 3 } }\left( { x }^{ 2 }+3 \right)$

"y"(x)>0" if $x\in [0,1)$ , "y"(x)<0" if $x\in(1,+\infty)$ . Hence function is convex down in $(0,1)$ and convex up in $(1,+\infty)$ .