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What is the domain of the function $f(x)=sqrt[3]x^3-x$?

5

$begingroup$

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

share|cite|improve this question

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$x^3-xgeq 0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Apr 14 at 9:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^frac13$ so $D(g)=x^3-x>0.$ All these a definition only.
  $endgroup$
  – Michael Rozenberg
  Apr 14 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner can you explain why?
  $endgroup$
  – J.Moh
  Apr 14 at 9:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$g(0)=0$$ dear Michael.
  $endgroup$
  – Dr. Sonnhard Graubner
  Apr 14 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you quite sure it wasn't $sqrtx^3-x$? Because the domain of $sqrt[3]x^3 -x$ is $Bbb R$.
  $endgroup$
  – TonyK
  Apr 14 at 18:21
  

  
  
  
  

 | 
show 1 more comment

5

$begingroup$

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

share|cite|improve this question

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$x^3-xgeq 0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Apr 14 at 9:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it's better $D(f)=mathbb R$, but if $g(x)=(x^3-x)^frac13$ so $D(g)=x^3-x>0.$ All these a definition only.
  $endgroup$
  – Michael Rozenberg
  Apr 14 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner can you explain why?
  $endgroup$
  – J.Moh
  Apr 14 at 9:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$g(0)=0$$ dear Michael.
  $endgroup$
  – Dr. Sonnhard Graubner
  Apr 14 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you quite sure it wasn't $sqrtx^3-x$? Because the domain of $sqrt[3]x^3 -x$ is $Bbb R$.
  $endgroup$
  – TonyK
  Apr 14 at 18:21
  

  
  
  
  

 | 
show 1 more 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

share|cite|improve this question

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

$endgroup$

share|cite|improve this question

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

share|cite|improve this question

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

edited Apr 14 at 18:18

Asaf Karagila♦

309k33441775

asked Apr 14 at 9:36

J.MohJ.Moh

695

asked Apr 14 at 9:36

J.MohJ.Moh

695

1 Answer
1

active

oldest

votes

12

$begingroup$

If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]phantomX$ 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).

share|cite|improve this answer

edited Apr 14 at 10:03

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
  $endgroup$
  – J.Moh
  Apr 14 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  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 $(mathbbR x ↦ x^n)$ is a bijection from $mathbbR$ to $mathbbR$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
  $endgroup$
  – user21820
  Apr 14 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
  $endgroup$
  – user21820
  Apr 14 at 15:32
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 14 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user21820 I did
  $endgroup$
  – J.Moh
  Apr 14 at 15:36
  

  
  
  
  

 | 
show 2 more comments

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12

$begingroup$

If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]phantomX$ 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).

share|cite|improve this answer

edited Apr 14 at 10:03

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
  $endgroup$
  – J.Moh
  Apr 14 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  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 $(mathbbR x ↦ x^n)$ is a bijection from $mathbbR$ to $mathbbR$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
  $endgroup$
  – user21820
  Apr 14 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
  $endgroup$
  – user21820
  Apr 14 at 15:32
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 14 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user21820 I did
  $endgroup$
  – J.Moh
  Apr 14 at 15:36
  

  
  
  
  

 | 
show 2 more comments

12

$begingroup$

If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]phantomX$ 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).

share|cite|improve this answer

edited Apr 14 at 10:03

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That s why I love computer scientists, they answer as if they re writing code ;) Thanks Henning! Perfect!
  $endgroup$
  – J.Moh
  Apr 14 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  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 $(mathbbR x ↦ x^n)$ is a bijection from $mathbbR$ to $mathbbR$, so its inverse exists. Similarly for non-negative real $x$ and even natural number $n$.
  $endgroup$
  – user21820
  Apr 14 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.Moh: By the way, if you are satisfied with this answer, you can click the tick to accept it.
  $endgroup$
  – user21820
  Apr 14 at 15:32
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 14 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user21820 I did
  $endgroup$
  – J.Moh
  Apr 14 at 15:36
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

If your book reaches the domain $[-1,0]cup[1,+infty)$, it must be because the book only considers $sqrt[3]phantomX$ 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).

share|cite|improve this answer

edited Apr 14 at 10:03

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

$endgroup$

share|cite|improve this answer

edited Apr 14 at 10:03

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556

answered Apr 14 at 9:57

Henning MakholmHenning Makholm

244k17312556