What is the domain and range of function: $f(x) = arccos( -x^2 - frac 12)$?
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
edited Nov 27 '18 at 16:50
gimusi
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
add a comment |
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
edited Nov 27 '18 at 16:50
gimusi
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
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.
– gimusi
Nov 27 '18 at 16:51
add a comment |
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
edited Nov 27 '18 at 16:50
gimusi
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
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
edited Nov 27 '18 at 16:50
gimusi
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
edited Nov 27 '18 at 16:50
gimusi
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
edited Nov 27 '18 at 16:50
gimusi
1
edited Nov 27 '18 at 16:50
gimusi
1
edited Nov 27 '18 at 16:50
gimusi
1
1
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
asked Nov 27 '18 at 10:54
Sergey VorobyevSergey Vorobyev
122
122
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.
– gimusi
Nov 27 '18 at 16:51
add a comment |
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.
– gimusi
Nov 27 '18 at 16:51
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.
– gimusi
Nov 27 '18 at 16:51
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.
– gimusi
Nov 27 '18 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
HINT
Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.
What are domain and range for $g(x)=-x^2-frac12$?
answered Nov 27 '18 at 10:57
gimusigimusi
1
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
|
show 5 more comments
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).
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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HINT
Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.
What are domain and range for $g(x)=-x^2-frac12$?
answered Nov 27 '18 at 10:57
gimusigimusi
1
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
|
show 5 more comments
HINT
Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.
What are domain and range for $g(x)=-x^2-frac12$?
answered Nov 27 '18 at 10:57
gimusigimusi
1
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
|
show 5 more comments
HINT
Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.
What are domain and range for $g(x)=-x^2-frac12$?
answered Nov 27 '18 at 10:57
gimusigimusi
1
HINT
Recall that $arccos (y)$ has domain $[-1,1]$ and range $[0,pi]$.
What are domain and range for $g(x)=-x^2-frac12$?
answered Nov 27 '18 at 10:57
gimusigimusi
1
answered Nov 27 '18 at 10:57
gimusigimusi
1
answered Nov 27 '18 at 10:57
gimusigimusi
1
answered Nov 27 '18 at 10:57
gimusigimusi
1
1
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
|
show 5 more comments
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
Domain for g(x) = [-1/(√2, 1/(√2)], and Range is all real numbers.
– Sergey Vorobyev
Nov 27 '18 at 12:41
@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?
– gimusi
Nov 27 '18 at 12:48
@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?
– gimusi
Nov 27 '18 at 12:48
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
Ah, yeah, I see it.
– Sergey Vorobyev
Nov 27 '18 at 13:05
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev That's nice! Now you need to make a restriction for g(x) such that its range fits with the domain for $arccos((g(x))$.
– gimusi
Nov 27 '18 at 13:06
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
@SergeyVorobyev If you show your work editing your question, I'll take a look to it. Bye
– gimusi
Nov 27 '18 at 13:08
|
show 5 more comments
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).
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
add a comment |
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).
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
add a comment |
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).
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
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).
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
edited Dec 13 '18 at 17:15
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
answered Nov 27 '18 at 16:20
Sergey VorobyevSergey Vorobyev
122
122
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
add a comment |
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
Let add your derivation directly in your original question by editing it.
– gimusi
Nov 27 '18 at 16:40
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
I don't know any about it yet, sorry. :-)
– Sergey Vorobyev
Dec 4 '18 at 9:21
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
Try to follow the suggestion given as a comment under your OP. Try to revise that accordinlgy. You can delete that answer.
– gimusi
Dec 4 '18 at 9:54
add a comment |
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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.
– gimusi
Nov 27 '18 at 16:51