What is the domain and range of function: $f(x) = arccos( -x^2 - frac 12)$?












-1












$begingroup$


What is the domain and range of function: $f(x) = arccos( -x^2 - frac12)$ ?



My attempt:



We have $f(x)=arccos(-x^2- 1/2)$, $g(x)= -x^2- 1/2$, $-1 le -x^2 - 1/2 le -1$ and $x^2 ge 0$.



So $x^2 le1/2$.



According that the domain of $g(x)$ is $Dg = [-1/√2 , 1/√2 ]$.



And the Range of $g(x)$ is $Rg = (-∞ , -1/2)$.



So the Range of g(x) will be our Domain of our function $f(x) ( f(x) = arccos(g(x)))$ will be $Df = (-∞, -1/2)$.



In this case, the Range of our function $f(x)$ will be $Rf = (2π/3 , π)$.






analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:51
















-1












$begingroup$


What is the domain and range of function: $f(x) = arccos( -x^2 - frac12)$ ?



My attempt:



We have $f(x)=arccos(-x^2- 1/2)$, $g(x)= -x^2- 1/2$, $-1 le -x^2 - 1/2 le -1$ and $x^2 ge 0$.



So $x^2 le1/2$.



According that the domain of $g(x)$ is $Dg = [-1/√2 , 1/√2 ]$.



And the Range of $g(x)$ is $Rg = (-∞ , -1/2)$.



So the Range of g(x) will be our Domain of our function $f(x) ( f(x) = arccos(g(x)))$ will be $Df = (-∞, -1/2)$.



In this case, the Range of our function $f(x)$ will be $Rf = (2π/3 , π)$.






analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:51














-1












-1








-1





$begingroup$


What is the domain and range of function: $f(x) = arccos( -x^2 - frac12)$ ?



My attempt:



We have $f(x)=arccos(-x^2- 1/2)$, $g(x)= -x^2- 1/2$, $-1 le -x^2 - 1/2 le -1$ and $x^2 ge 0$.



So $x^2 le1/2$.



According that the domain of $g(x)$ is $Dg = [-1/√2 , 1/√2 ]$.



And the Range of $g(x)$ is $Rg = (-∞ , -1/2)$.



So the Range of g(x) will be our Domain of our function $f(x) ( f(x) = arccos(g(x)))$ will be $Df = (-∞, -1/2)$.



In this case, the Range of our function $f(x)$ will be $Rf = (2π/3 , π)$.






analysis







share|cite|improve this question











$endgroup$




What is the domain and range of function: $f(x) = arccos( -x^2 - frac12)$ ?



My attempt:



We have $f(x)=arccos(-x^2- 1/2)$, $g(x)= -x^2- 1/2$, $-1 le -x^2 - 1/2 le -1$ and $x^2 ge 0$.



So $x^2 le1/2$.



According that the domain of $g(x)$ is $Dg = [-1/√2 , 1/√2 ]$.



And the Range of $g(x)$ is $Rg = (-∞ , -1/2)$.



So the Range of g(x) will be our Domain of our function $f(x) ( f(x) = arccos(g(x)))$ will be $Df = (-∞, -1/2)$.



In this case, the Range of our function $f(x)$ will be $Rf = (2π/3 , π)$.






analysis




analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 16:50









gimusi

92.8k84494




92.8k84494










asked Nov 27 '18 at 10:54









Sergey VorobyevSergey Vorobyev

122




122












  • $begingroup$
    By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:51


















  • $begingroup$
    By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:51
















$begingroup$
By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
$endgroup$
– gimusi
Nov 27 '18 at 16:51




$begingroup$
By the restriction to $Dg = [-1/√2 , 1/√2 ]$ the range of g(x) is $Rg = (-1,-1/2)$ and then the range for f(x) is $Rf = (2π/3 , π)$ as you correctly stated. Therefore the only problem is with the range for $g(x)$ which you have indicated.
$endgroup$
– gimusi
Nov 27 '18 at 16:51










2 Answers
2






active

oldest

votes


















3












$begingroup$

HINT



Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.



What are domain and range for $g(x)=-x^2-frac12$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 12:41










  • $begingroup$
    @SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
    $endgroup$
    – gimusi
    Nov 27 '18 at 12:48










  • $begingroup$
    Ah, yeah, I see it.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 13:05










  • $begingroup$
    @SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:06










  • $begingroup$
    @SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:08



















0












$begingroup$

f(x)=arccos(-x^2- 1/2). g(x)= -x^2- 1/2. -1 <= -x^2 - 1/2 <= 1 and x^2 >= 0. So x^2 <=1/2. According that the domain of g(x) is Dg = [-1/√2 , 1/√2 ]. And the Range of g(x) is Rg = (-∞ , -1/2). So the Range of g(x) will be our Domain of our function f(x) ( f(x) = arccos(g(x))) will be Df = (-∞, -1/2). In this case, the Range of our function f(x) will beRf = (2π/3 , π).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Let add your derivation directly in your original question by editing it.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:40










  • $begingroup$
    I don't know any about it yet, sorry. :-)
    $endgroup$
    – Sergey Vorobyev
    Dec 4 '18 at 9:21










  • $begingroup$
    Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
    $endgroup$
    – gimusi
    Dec 4 '18 at 9:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

HINT



Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.



What are domain and range for $g(x)=-x^2-frac12$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 12:41










  • $begingroup$
    @SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
    $endgroup$
    – gimusi
    Nov 27 '18 at 12:48










  • $begingroup$
    Ah, yeah, I see it.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 13:05










  • $begingroup$
    @SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:06










  • $begingroup$
    @SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:08
















3












$begingroup$

HINT



Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.



What are domain and range for $g(x)=-x^2-frac12$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 12:41










  • $begingroup$
    @SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
    $endgroup$
    – gimusi
    Nov 27 '18 at 12:48










  • $begingroup$
    Ah, yeah, I see it.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 13:05










  • $begingroup$
    @SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:06










  • $begingroup$
    @SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:08














3












3








3





$begingroup$

HINT



Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.



What are domain and range for $g(x)=-x^2-frac12$?






share|cite|improve this answer









$endgroup$



HINT



Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.



What are domain and range for $g(x)=-x^2-frac12$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 10:57









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 12:41










  • $begingroup$
    @SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
    $endgroup$
    – gimusi
    Nov 27 '18 at 12:48










  • $begingroup$
    Ah, yeah, I see it.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 13:05










  • $begingroup$
    @SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:06










  • $begingroup$
    @SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:08


















  • $begingroup$
    Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 12:41










  • $begingroup$
    @SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
    $endgroup$
    – gimusi
    Nov 27 '18 at 12:48










  • $begingroup$
    Ah, yeah, I see it.
    $endgroup$
    – Sergey Vorobyev
    Nov 27 '18 at 13:05










  • $begingroup$
    @SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:06










  • $begingroup$
    @SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
    $endgroup$
    – gimusi
    Nov 27 '18 at 13:08
















$begingroup$
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
$endgroup$
– Sergey Vorobyev
Nov 27 '18 at 12:41




$begingroup$
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
$endgroup$
– Sergey Vorobyev
Nov 27 '18 at 12:41












$begingroup$
@SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
$endgroup$
– gimusi
Nov 27 '18 at 12:48




$begingroup$
@SergeyVorobyev Recall that the domain for $g(x)$ is the set of all values where $g(x)$ is defined that is all $mathbb{R}$ and the range is $(-infty,-1/2]$. Can you see why?
$endgroup$
– gimusi
Nov 27 '18 at 12:48












$begingroup$
Ah, yeah, I see it.
$endgroup$
– Sergey Vorobyev
Nov 27 '18 at 13:05




$begingroup$
Ah, yeah, I see it.
$endgroup$
– Sergey Vorobyev
Nov 27 '18 at 13:05












$begingroup$
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
$endgroup$
– gimusi
Nov 27 '18 at 13:06




$begingroup$
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
$endgroup$
– gimusi
Nov 27 '18 at 13:06












$begingroup$
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
$endgroup$
– gimusi
Nov 27 '18 at 13:08




$begingroup$
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
$endgroup$
– gimusi
Nov 27 '18 at 13:08











0












$begingroup$

f(x)=arccos(-x^2- 1/2). g(x)= -x^2- 1/2. -1 <= -x^2 - 1/2 <= 1 and x^2 >= 0. So x^2 <=1/2. According that the domain of g(x) is Dg = [-1/√2 , 1/√2 ]. And the Range of g(x) is Rg = (-∞ , -1/2). So the Range of g(x) will be our Domain of our function f(x) ( f(x) = arccos(g(x))) will be Df = (-∞, -1/2). In this case, the Range of our function f(x) will beRf = (2π/3 , π).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Let add your derivation directly in your original question by editing it.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:40










  • $begingroup$
    I don't know any about it yet, sorry. :-)
    $endgroup$
    – Sergey Vorobyev
    Dec 4 '18 at 9:21










  • $begingroup$
    Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
    $endgroup$
    – gimusi
    Dec 4 '18 at 9:54
















0












$begingroup$

f(x)=arccos(-x^2- 1/2). g(x)= -x^2- 1/2. -1 <= -x^2 - 1/2 <= 1 and x^2 >= 0. So x^2 <=1/2. According that the domain of g(x) is Dg = [-1/√2 , 1/√2 ]. And the Range of g(x) is Rg = (-∞ , -1/2). So the Range of g(x) will be our Domain of our function f(x) ( f(x) = arccos(g(x))) will be Df = (-∞, -1/2). In this case, the Range of our function f(x) will beRf = (2π/3 , π).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Let add your derivation directly in your original question by editing it.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:40










  • $begingroup$
    I don't know any about it yet, sorry. :-)
    $endgroup$
    – Sergey Vorobyev
    Dec 4 '18 at 9:21










  • $begingroup$
    Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
    $endgroup$
    – gimusi
    Dec 4 '18 at 9:54














0












0








0





$begingroup$

f(x)=arccos(-x^2- 1/2). g(x)= -x^2- 1/2. -1 <= -x^2 - 1/2 <= 1 and x^2 >= 0. So x^2 <=1/2. According that the domain of g(x) is Dg = [-1/√2 , 1/√2 ]. And the Range of g(x) is Rg = (-∞ , -1/2). So the Range of g(x) will be our Domain of our function f(x) ( f(x) = arccos(g(x))) will be Df = (-∞, -1/2). In this case, the Range of our function f(x) will beRf = (2π/3 , π).






share|cite|improve this answer











$endgroup$



f(x)=arccos(-x^2- 1/2). g(x)= -x^2- 1/2. -1 <= -x^2 - 1/2 <= 1 and x^2 >= 0. So x^2 <=1/2. According that the domain of g(x) is Dg = [-1/√2 , 1/√2 ]. And the Range of g(x) is Rg = (-∞ , -1/2). So the Range of g(x) will be our Domain of our function f(x) ( f(x) = arccos(g(x))) will be Df = (-∞, -1/2). In this case, the Range of our function f(x) will beRf = (2π/3 , π).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 17:15

























answered Nov 27 '18 at 16:20









Sergey VorobyevSergey Vorobyev

122




122












  • $begingroup$
    Let add your derivation directly in your original question by editing it.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:40










  • $begingroup$
    I don't know any about it yet, sorry. :-)
    $endgroup$
    – Sergey Vorobyev
    Dec 4 '18 at 9:21










  • $begingroup$
    Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
    $endgroup$
    – gimusi
    Dec 4 '18 at 9:54


















  • $begingroup$
    Let add your derivation directly in your original question by editing it.
    $endgroup$
    – gimusi
    Nov 27 '18 at 16:40










  • $begingroup$
    I don't know any about it yet, sorry. :-)
    $endgroup$
    – Sergey Vorobyev
    Dec 4 '18 at 9:21










  • $begingroup$
    Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
    $endgroup$
    – gimusi
    Dec 4 '18 at 9:54
















$begingroup$
Let add your derivation directly in your original question by editing it.
$endgroup$
– gimusi
Nov 27 '18 at 16:40




$begingroup$
Let add your derivation directly in your original question by editing it.
$endgroup$
– gimusi
Nov 27 '18 at 16:40












$begingroup$
I don't know any about it yet, sorry. :-)
$endgroup$
– Sergey Vorobyev
Dec 4 '18 at 9:21




$begingroup$
I don't know any about it yet, sorry. :-)
$endgroup$
– Sergey Vorobyev
Dec 4 '18 at 9:21












$begingroup$
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
$endgroup$
– gimusi
Dec 4 '18 at 9:54




$begingroup$
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
$endgroup$
– gimusi
Dec 4 '18 at 9:54


















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