What is the domain of the function $f(x)=sqrt[3]{x^3-x}$?
$begingroup$
Let $f$ be: $f(x) = sqrt[3]{x^3 -x}$, an exercise book asked for the domain of definition. Isn't it over $mathbb R$. The book solution stated $Df = [-1,0] cup [1, +infty[$
I don t get it. Can you explain?
calculus
edited 9 hours ago
Asaf Karagila♦
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
$endgroup$
add a comment |
$begingroup$
Let $f$ be: $f(x) = sqrt[3]{x^3 -x}$, an exercise book asked for the domain of definition. Isn't it over $mathbb R$. The book solution stated $Df = [-1,0] cup [1, +infty[$
I don t get it. Can you explain?
calculus
edited 9 hours ago
Asaf Karagila♦
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
$endgroup$
$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago
add a comment |
$begingroup$
Let $f$ be: $f(x) = sqrt[3]{x^3 -x}$, an exercise book asked for the domain of definition. Isn't it over $mathbb R$. The book solution stated $Df = [-1,0] cup [1, +infty[$
I don t get it. Can you explain?
calculus
edited 9 hours ago
Asaf Karagila♦
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
$endgroup$
Let $f$ be: $f(x) = sqrt[3]{x^3 -x}$, an exercise book asked for the domain of definition. Isn't it over $mathbb R$. The book solution stated $Df = [-1,0] cup [1, +infty[$
I don t get it. Can you explain?
calculus
calculus
edited 9 hours ago
Asaf Karagila♦
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
edited 9 hours ago
Asaf Karagila♦
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
edited 9 hours ago
Asaf Karagila♦
308k33441775
edited 9 hours ago
Asaf Karagila♦
308k33441775
edited 9 hours ago
Asaf Karagila♦
308k33441775
308k33441775
asked 17 hours ago
J.MohJ.Moh
855
asked 17 hours ago
J.MohJ.Moh
855
asked 17 hours ago
J.MohJ.Moh
855
855
$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago
add a comment |
$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago
$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]{phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $sqrt[N]{phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $xmapsto x^N$ is a bijection on $mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]{phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $sqrt[N]{phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $xmapsto x^N$ is a bijection on $mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
|
show 2 more comments
$begingroup$
If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]{phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $sqrt[N]{phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $xmapsto x^N$ is a bijection on $mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
$endgroup$
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
|
show 2 more comments
$begingroup$
If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]{phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $sqrt[N]{phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $xmapsto x^N$ is a bijection on $mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
$endgroup$
If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]{phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $sqrt[N]{phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $xmapsto x^N$ is a bijection on $mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
edited 17 hours ago
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
answered 17 hours ago
Henning MakholmHenning Makholm
243k17312556
243k17312556
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
|
show 2 more comments
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
$begingroup$
That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
$endgroup$
– J.Moh
17 hours ago
1
1
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: If every mathematics student learns programming, we would hardly see any of the silly mistakes arising from imprecision. I agree with Henning's last sentence and even say that such kind of questions are terrible because they encourage imprecision. Moreover, I personally think that we should define $sqrt[n]{x}$ for all real $x$ and odd natural number $n$, because $(mathbb{R} x ↦ x^n)$ is a bijection from $mathbb{R}$ to $mathbb{R}$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
$begingroup$
@J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
$endgroup$
– user21820
11 hours ago
1
1
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
I agree! I found refuge in math and programming since I could sense for the first time what honesty was.
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
$begingroup$
@user21820 I did
$endgroup$
– J.Moh
11 hours ago
|
show 2 more comments
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$begingroup$
It is $$x^3-xgeq 0$$
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^{frac{1}{3}}$ so $D(g)={xinmathbb R|x^3-x>0}.$ All these a definition only.
$endgroup$
– Michael Rozenberg
17 hours ago
$begingroup$
@Dr.SonnhardGraubner can you explain why?
$endgroup$
– J.Moh
17 hours ago
$begingroup$
$$g(0)=0$$ dear Michael.
$endgroup$
– Dr. Sonnhard Graubner
17 hours ago
$begingroup$
Are you quite sure it wasn't $sqrt{x^3-x}$? Because the domain of $sqrt[3]{x^3 -x}$ is $Bbb R$.
$endgroup$
– TonyK
9 hours ago