Set of $f^-1((-2,5))$ [on hold]

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up vote
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Considering the function $f: mathbbRrightarrowmathbbR$ defined by $f(x) = x^2$ the set defined by $f^-1((-2,5))$ is? (For improvement)

elementary-functions

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edited 9 hours ago

asked 12 hours ago

Lamech Muluya Kibudde

113

put on hold as off-topic by Shaun, Arnaud Mortier, amWhy, callculus, Key Flex 11 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Arnaud Mortier, amWhy, callculus, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What have you tried?
  – alphacapture
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Abstract algebra and measure theory?
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @alphacapture Please read this.
  – Shaun
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
  – Lamech Muluya Kibudde
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
  – Arnaud Mortier
  11 hours ago
  

  
  
  
  

 | 
show 2 more comments

up vote
-6
down vote

favorite

Considering the function $f: mathbbRrightarrowmathbbR$ defined by $f(x) = x^2$ the set defined by $f^-1((-2,5))$ is? (For improvement)

elementary-functions

share|cite|improve this question

edited 9 hours ago

asked 12 hours ago

Lamech Muluya Kibudde

113

put on hold as off-topic by Shaun, Arnaud Mortier, amWhy, callculus, Key Flex 11 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Arnaud Mortier, amWhy, callculus, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What have you tried?
  – alphacapture
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Abstract algebra and measure theory?
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @alphacapture Please read this.
  – Shaun
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
  – Lamech Muluya Kibudde
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
  – Arnaud Mortier
  11 hours ago
  

  
  
  
  

 | 
show 2 more comments

up vote
-6
down vote

favorite

up vote
-6
down vote

favorite

Considering the function $f: mathbbRrightarrowmathbbR$ defined by $f(x) = x^2$ the set defined by $f^-1((-2,5))$ is? (For improvement)

elementary-functions

share|cite|improve this question

edited 9 hours ago

asked 12 hours ago

Lamech Muluya Kibudde

113

Considering the function $f: mathbbRrightarrowmathbbR$ defined by $f(x) = x^2$ the set defined by $f^-1((-2,5))$ is? (For improvement)

elementary-functions

share|cite|improve this question

edited 9 hours ago

asked 12 hours ago

Lamech Muluya Kibudde

113

share|cite|improve this question

share|cite|improve this question

edited 9 hours ago

edited 9 hours ago

edited 9 hours ago

asked 12 hours ago

Lamech Muluya Kibudde

113

asked 12 hours ago

Lamech Muluya Kibudde

113

asked 12 hours ago

Lamech Muluya Kibudde

113

113

put on hold as off-topic by Shaun, Arnaud Mortier, amWhy, callculus, Key Flex 11 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Arnaud Mortier, amWhy, callculus, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Shaun, Arnaud Mortier, amWhy, callculus, Key Flex 11 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Arnaud Mortier, amWhy, callculus, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What have you tried?
  – alphacapture
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Abstract algebra and measure theory?
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @alphacapture Please read this.
  – Shaun
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
  – Lamech Muluya Kibudde
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
  – Arnaud Mortier
  11 hours ago
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What have you tried?
  – alphacapture
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Abstract algebra and measure theory?
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @alphacapture Please read this.
  – Shaun
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
  – Lamech Muluya Kibudde
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
  – Arnaud Mortier
  11 hours ago
  

  
  
  
  

1

1

What have you tried?
– alphacapture
12 hours ago

What have you tried?
– alphacapture
12 hours ago

Abstract algebra and measure theory?
– Arnaud Mortier
12 hours ago

Abstract algebra and measure theory?
– Arnaud Mortier
12 hours ago

@alphacapture Please read this.
– Shaun
12 hours ago

@alphacapture Please read this.
– Shaun
12 hours ago

My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
– Lamech Muluya Kibudde
12 hours ago

My expectation would be to get a set of elements e.g (Sqrt -5, Sqrt 5) Just dont know how to get it,
– Lamech Muluya Kibudde
12 hours ago

1

1

@janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
– Arnaud Mortier
11 hours ago

@janmarqz Before you make such an edit that will completely change the meaning of the question, please ask the OP clarification about what they mean.
– Arnaud Mortier
11 hours ago

 | 
show 2 more comments

2 Answers
2

active

oldest

votes

up vote
2
down vote

Let $f : mathbbR rightarrow mathbbR$ be a map defined by $f(x) = x^2$. Then $f^-1((-2,5)) = lbrace x in mathbbR : f(x) in (-2,5) rbrace$. Now, since our function only assumes positives values, we are essentially finding those $f(x) in mathbbR$ such that $x in [0,5)$. Hence $f^-1((-2,5)) = (-sqrt5, sqrt5)$.

share|cite|improve this answer

edited 11 hours ago

community wiki

2 revs
Fabian

add a comment | 

up vote
1
down vote

Recall that $x^2geq 0;;forall x$. Then we can divide the set $(-2,5)$ as two sets, one with the numbers in that set less then zero and one with the greater or equal to zero $$(-2,5) = (-2,0)cup[0,5)$$ Being the function $f(x)=x^2$ always postive, the first set cannot be part of the image of $f$, so $$f^-1((-2,0))=emptyset$$ on the other hand, the other subset has preimage. You can easily see that by inverting the function $$f^-1(x) = pmsqrtx$$ you find that $$f^-1(0)=0;;;f^-1(5) = pmsqrt5$$ because all for all the other values in the subset $[0,5)$ the function $f^-1$ is defined you get that $$f^-1([0,5)) = (-sqrt5,sqrt5)$$ At the end, being the other set empty set, you get $$f^-1((-2,5)) = (-sqrt5,sqrt5)$$

Maybe next time you should give us your attempts. The only way of learning mathematics is by studying and lots of trial and errors. Don't be ashamed if you cannot get the right answer right away, that's part for the course. See, even I made an error that led me to a totally different answer from the right one! But by carefully checking the calculations helps every time

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is quite helpful Davide
  – Lamech Muluya Kibudde
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
  – Davide Morgante
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes. You have forgotten the $pm$ sign in the square root.
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 Oh ok, you're right! Thank you
  – Davide Morgante
  11 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Let $f : mathbbR rightarrow mathbbR$ be a map defined by $f(x) = x^2$. Then $f^-1((-2,5)) = lbrace x in mathbbR : f(x) in (-2,5) rbrace$. Now, since our function only assumes positives values, we are essentially finding those $f(x) in mathbbR$ such that $x in [0,5)$. Hence $f^-1((-2,5)) = (-sqrt5, sqrt5)$.

share|cite|improve this answer

edited 11 hours ago

community wiki

2 revs
Fabian

add a comment | 

up vote
2
down vote

Let $f : mathbbR rightarrow mathbbR$ be a map defined by $f(x) = x^2$. Then $f^-1((-2,5)) = lbrace x in mathbbR : f(x) in (-2,5) rbrace$. Now, since our function only assumes positives values, we are essentially finding those $f(x) in mathbbR$ such that $x in [0,5)$. Hence $f^-1((-2,5)) = (-sqrt5, sqrt5)$.

share|cite|improve this answer

edited 11 hours ago

community wiki

2 revs
Fabian

add a comment | 

up vote
2
down vote

up vote
2
down vote

Let $f : mathbbR rightarrow mathbbR$ be a map defined by $f(x) = x^2$. Then $f^-1((-2,5)) = lbrace x in mathbbR : f(x) in (-2,5) rbrace$. Now, since our function only assumes positives values, we are essentially finding those $f(x) in mathbbR$ such that $x in [0,5)$. Hence $f^-1((-2,5)) = (-sqrt5, sqrt5)$.

share|cite|improve this answer

edited 11 hours ago

community wiki

2 revs
Fabian

Let $f : mathbbR rightarrow mathbbR$ be a map defined by $f(x) = x^2$. Then $f^-1((-2,5)) = lbrace x in mathbbR : f(x) in (-2,5) rbrace$. Now, since our function only assumes positives values, we are essentially finding those $f(x) in mathbbR$ such that $x in [0,5)$. Hence $f^-1((-2,5)) = (-sqrt5, sqrt5)$.

share|cite|improve this answer

edited 11 hours ago

community wiki

2 revs
Fabian

share|cite|improve this answer

share|cite|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

community wiki

2 revs
Fabian

community wiki

2 revs
Fabian

community wiki

2 revs
Fabian

add a comment | 

add a comment | 

up vote
1
down vote

Recall that $x^2geq 0;;forall x$. Then we can divide the set $(-2,5)$ as two sets, one with the numbers in that set less then zero and one with the greater or equal to zero $$(-2,5) = (-2,0)cup[0,5)$$ Being the function $f(x)=x^2$ always postive, the first set cannot be part of the image of $f$, so $$f^-1((-2,0))=emptyset$$ on the other hand, the other subset has preimage. You can easily see that by inverting the function $$f^-1(x) = pmsqrtx$$ you find that $$f^-1(0)=0;;;f^-1(5) = pmsqrt5$$ because all for all the other values in the subset $[0,5)$ the function $f^-1$ is defined you get that $$f^-1([0,5)) = (-sqrt5,sqrt5)$$ At the end, being the other set empty set, you get $$f^-1((-2,5)) = (-sqrt5,sqrt5)$$

Maybe next time you should give us your attempts. The only way of learning mathematics is by studying and lots of trial and errors. Don't be ashamed if you cannot get the right answer right away, that's part for the course. See, even I made an error that led me to a totally different answer from the right one! But by carefully checking the calculations helps every time

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is quite helpful Davide
  – Lamech Muluya Kibudde
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
  – Davide Morgante
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes. You have forgotten the $pm$ sign in the square root.
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 Oh ok, you're right! Thank you
  – Davide Morgante
  11 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

Recall that $x^2geq 0;;forall x$. Then we can divide the set $(-2,5)$ as two sets, one with the numbers in that set less then zero and one with the greater or equal to zero $$(-2,5) = (-2,0)cup[0,5)$$ Being the function $f(x)=x^2$ always postive, the first set cannot be part of the image of $f$, so $$f^-1((-2,0))=emptyset$$ on the other hand, the other subset has preimage. You can easily see that by inverting the function $$f^-1(x) = pmsqrtx$$ you find that $$f^-1(0)=0;;;f^-1(5) = pmsqrt5$$ because all for all the other values in the subset $[0,5)$ the function $f^-1$ is defined you get that $$f^-1([0,5)) = (-sqrt5,sqrt5)$$ At the end, being the other set empty set, you get $$f^-1((-2,5)) = (-sqrt5,sqrt5)$$

Maybe next time you should give us your attempts. The only way of learning mathematics is by studying and lots of trial and errors. Don't be ashamed if you cannot get the right answer right away, that's part for the course. See, even I made an error that led me to a totally different answer from the right one! But by carefully checking the calculations helps every time

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is quite helpful Davide
  – Lamech Muluya Kibudde
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
  – Davide Morgante
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes. You have forgotten the $pm$ sign in the square root.
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 Oh ok, you're right! Thank you
  – Davide Morgante
  11 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

up vote
1
down vote

Recall that $x^2geq 0;;forall x$. Then we can divide the set $(-2,5)$ as two sets, one with the numbers in that set less then zero and one with the greater or equal to zero $$(-2,5) = (-2,0)cup[0,5)$$ Being the function $f(x)=x^2$ always postive, the first set cannot be part of the image of $f$, so $$f^-1((-2,0))=emptyset$$ on the other hand, the other subset has preimage. You can easily see that by inverting the function $$f^-1(x) = pmsqrtx$$ you find that $$f^-1(0)=0;;;f^-1(5) = pmsqrt5$$ because all for all the other values in the subset $[0,5)$ the function $f^-1$ is defined you get that $$f^-1([0,5)) = (-sqrt5,sqrt5)$$ At the end, being the other set empty set, you get $$f^-1((-2,5)) = (-sqrt5,sqrt5)$$

Maybe next time you should give us your attempts. The only way of learning mathematics is by studying and lots of trial and errors. Don't be ashamed if you cannot get the right answer right away, that's part for the course. See, even I made an error that led me to a totally different answer from the right one! But by carefully checking the calculations helps every time

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

Recall that $x^2geq 0;;forall x$. Then we can divide the set $(-2,5)$ as two sets, one with the numbers in that set less then zero and one with the greater or equal to zero $$(-2,5) = (-2,0)cup[0,5)$$ Being the function $f(x)=x^2$ always postive, the first set cannot be part of the image of $f$, so $$f^-1((-2,0))=emptyset$$ on the other hand, the other subset has preimage. You can easily see that by inverting the function $$f^-1(x) = pmsqrtx$$ you find that $$f^-1(0)=0;;;f^-1(5) = pmsqrt5$$ because all for all the other values in the subset $[0,5)$ the function $f^-1$ is defined you get that $$f^-1([0,5)) = (-sqrt5,sqrt5)$$ At the end, being the other set empty set, you get $$f^-1((-2,5)) = (-sqrt5,sqrt5)$$

Maybe next time you should give us your attempts. The only way of learning mathematics is by studying and lots of trial and errors. Don't be ashamed if you cannot get the right answer right away, that's part for the course. See, even I made an error that led me to a totally different answer from the right one! But by carefully checking the calculations helps every time

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

share|cite|improve this answer

share|cite|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

answered 11 hours ago

Davide Morgante

1,574120

answered 11 hours ago

Davide Morgante

1,574120

answered 11 hours ago

Davide Morgante

1,574120

1,574120

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is quite helpful Davide
  – Lamech Muluya Kibudde
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
  – Davide Morgante
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes. You have forgotten the $pm$ sign in the square root.
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 Oh ok, you're right! Thank you
  – Davide Morgante
  11 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is quite helpful Davide
  – Lamech Muluya Kibudde
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
  – Davide Morgante
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes. You have forgotten the $pm$ sign in the square root.
  – Dog_69
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dog_69 Oh ok, you're right! Thank you
  – Davide Morgante
  11 hours ago
  
  

  
  
  
  

Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
– Dog_69
11 hours ago

Since $f$ is continuous, shouldn't $f^-1((-2,5))$ be open?
– Dog_69
11 hours ago

This is quite helpful Davide
– Lamech Muluya Kibudde
11 hours ago

This is quite helpful Davide
– Lamech Muluya Kibudde
11 hours ago

@Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
– Davide Morgante
11 hours ago

@Dog_69 mhh you're correct. Maybe there's something wrong in my calculations?
– Davide Morgante
11 hours ago

2

2

Yes. You have forgotten the $pm$ sign in the square root.
– Dog_69
11 hours ago

Yes. You have forgotten the $pm$ sign in the square root.
– Dog_69
11 hours ago

@Dog_69 Oh ok, you're right! Thank you
– Davide Morgante
11 hours ago

@Dog_69 Oh ok, you're right! Thank you
– Davide Morgante
11 hours ago

 | 
show 1 more comment