Is there inverse to right for this function?
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We have function $g: mathbb Rto[11/4, infty)$, with $g(x)=x^2-3x+5$ and we have to check if there exists $h:mathbb R to(-1, infty) $ with property $g circ h=1_{[11/4, infty)}$, where $circ$ means the composition of $g$ and $h$, and if it exists the problem ask to determine it.
I want to say that if $g$ is not a surjective function, that there not exist a inverse to right, but I don't know if it is totally correct, and from there how to find $h$?
functions inverse inverse-function
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
add a comment |
up vote
0
down vote
favorite
We have function $g: mathbb Rto[11/4, infty)$, with $g(x)=x^2-3x+5$ and we have to check if there exists $h:mathbb R to(-1, infty) $ with property $g circ h=1_{[11/4, infty)}$, where $circ$ means the composition of $g$ and $h$, and if it exists the problem ask to determine it.
I want to say that if $g$ is not a surjective function, that there not exist a inverse to right, but I don't know if it is totally correct, and from there how to find $h$?
functions inverse inverse-function
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have function $g: mathbb Rto[11/4, infty)$, with $g(x)=x^2-3x+5$ and we have to check if there exists $h:mathbb R to(-1, infty) $ with property $g circ h=1_{[11/4, infty)}$, where $circ$ means the composition of $g$ and $h$, and if it exists the problem ask to determine it.
I want to say that if $g$ is not a surjective function, that there not exist a inverse to right, but I don't know if it is totally correct, and from there how to find $h$?
functions inverse inverse-function
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
We have function $g: mathbb Rto[11/4, infty)$, with $g(x)=x^2-3x+5$ and we have to check if there exists $h:mathbb R to(-1, infty) $ with property $g circ h=1_{[11/4, infty)}$, where $circ$ means the composition of $g$ and $h$, and if it exists the problem ask to determine it.
I want to say that if $g$ is not a surjective function, that there not exist a inverse to right, but I don't know if it is totally correct, and from there how to find $h$?
functions inverse inverse-function
functions inverse inverse-function
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
edited Nov 27 at 11:57
N. F. Taussig
43.1k93254
43.1k93254
asked Nov 26 at 21:36
Raul1998
63
asked Nov 26 at 21:36
Raul1998
63
asked Nov 26 at 21:36
Raul1998
63
63
What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48
add a comment |
What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48
What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Hints:
$g$ is surjective because.
$g(x)=x^2-3x+5=left(x-frac{3}{2}right)^2+frac{11}{4}$. So the range of $g$ is $left[frac{11}{4},inftyright)$.
Since $g$ is surjective so $g$ does have a right inverse. To find that
begin{align*}
y & = x^2-3x+5\
x^2-3x+(5-y)&=0\
x & = frac{3pmsqrt{9-4(5-y)}}{2}\
x & = frac{3pmsqrt{4y-11}}{2}.
end{align*}
Now you have to choose the correct branch for the right inverse $h$. Can you proceed from here?
Also you need to define your $h$ separately for $Bbb{R}-left[frac{11}{4},inftyright)$.
answered Nov 26 at 21:49
Anurag A
25.4k12250
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hints:
$g$ is surjective because.
$g(x)=x^2-3x+5=left(x-frac{3}{2}right)^2+frac{11}{4}$. So the range of $g$ is $left[frac{11}{4},inftyright)$.
Since $g$ is surjective so $g$ does have a right inverse. To find that
begin{align*}
y & = x^2-3x+5\
x^2-3x+(5-y)&=0\
x & = frac{3pmsqrt{9-4(5-y)}}{2}\
x & = frac{3pmsqrt{4y-11}}{2}.
end{align*}
Now you have to choose the correct branch for the right inverse $h$. Can you proceed from here?
Also you need to define your $h$ separately for $Bbb{R}-left[frac{11}{4},inftyright)$.
answered Nov 26 at 21:49
Anurag A
25.4k12250
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
add a comment |
up vote
1
down vote
Hints:
$g$ is surjective because.
$g(x)=x^2-3x+5=left(x-frac{3}{2}right)^2+frac{11}{4}$. So the range of $g$ is $left[frac{11}{4},inftyright)$.
Since $g$ is surjective so $g$ does have a right inverse. To find that
begin{align*}
y & = x^2-3x+5\
x^2-3x+(5-y)&=0\
x & = frac{3pmsqrt{9-4(5-y)}}{2}\
x & = frac{3pmsqrt{4y-11}}{2}.
end{align*}
Now you have to choose the correct branch for the right inverse $h$. Can you proceed from here?
Also you need to define your $h$ separately for $Bbb{R}-left[frac{11}{4},inftyright)$.
answered Nov 26 at 21:49
Anurag A
25.4k12250
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
add a comment |
up vote
1
down vote
up vote
1
down vote
Hints:
$g$ is surjective because.
$g(x)=x^2-3x+5=left(x-frac{3}{2}right)^2+frac{11}{4}$. So the range of $g$ is $left[frac{11}{4},inftyright)$.
Since $g$ is surjective so $g$ does have a right inverse. To find that
begin{align*}
y & = x^2-3x+5\
x^2-3x+(5-y)&=0\
x & = frac{3pmsqrt{9-4(5-y)}}{2}\
x & = frac{3pmsqrt{4y-11}}{2}.
end{align*}
Now you have to choose the correct branch for the right inverse $h$. Can you proceed from here?
Also you need to define your $h$ separately for $Bbb{R}-left[frac{11}{4},inftyright)$.
answered Nov 26 at 21:49
Anurag A
25.4k12250
Hints:
$g$ is surjective because.
$g(x)=x^2-3x+5=left(x-frac{3}{2}right)^2+frac{11}{4}$. So the range of $g$ is $left[frac{11}{4},inftyright)$.
Since $g$ is surjective so $g$ does have a right inverse. To find that
begin{align*}
y & = x^2-3x+5\
x^2-3x+(5-y)&=0\
x & = frac{3pmsqrt{9-4(5-y)}}{2}\
x & = frac{3pmsqrt{4y-11}}{2}.
end{align*}
Now you have to choose the correct branch for the right inverse $h$. Can you proceed from here?
Also you need to define your $h$ separately for $Bbb{R}-left[frac{11}{4},inftyright)$.
answered Nov 26 at 21:49
Anurag A
25.4k12250
edited Nov 27 at 13:19
answered Nov 26 at 21:49
Anurag A
25.4k12250
answered Nov 26 at 21:49
Anurag A
25.4k12250
answered Nov 26 at 21:49
Anurag A
25.4k12250
25.4k12250
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
add a comment |
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
Yes, I can sole it now but I have two question if you can answer: Why define h for $Bbb{R}-left[frac{11}{4},inftyright)$? And second: if g was not a surjective function, we could draw the conclusion that he don't have a inverse to right ? Thanks!
– Raul1998
Nov 26 at 22:54
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
@Raul1998 the reason you have to define $h$ on that set is because the function we will get for $h$ from the expression I have in my answer gives you $h$ for only $x geq 11/4$. But the requirement you have in your question about the domain of $h$ is that it should be ALL real numbers. Regarding your second question, the answer is yes. If $g$ was not onto, then the right inverse wouldn't exist.
– Anurag A
Nov 27 at 6:11
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
You meant to write $g(x) = x^2 - 3x + 5 = left(x - frac{3}{2}right)^2 + frac{11}{4}$.
– N. F. Taussig
Nov 27 at 11:59
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
@N.F.Taussig Thanks for pointing out the typo.
– Anurag A
Nov 27 at 13:20
add a comment |
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What do ypu mean by $g circ h=1_{[11/4, infty)}$?
– gimusi
Nov 26 at 22:01
@gimusi This means that h is an inverse of g and 1_{[11/4, infty)} means identity function. $1_{A}(x) = x$ for each x in A. Or it if neutral element in a group of functions.
– Raul1998
Nov 26 at 22:48