Mixing
Find the range of the following function.
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Find the range of the following function
$$f(x)=(x-7)(x+9)over(x-7)(x+1)$$
I tried to factor the two expressions(the way it is presented) in the numerator and denominator.
I dont know to proceed further.
Thanks for any help.
functions
asked 52 mins ago
Love Invariants
93215
add a comment |Â
up vote
4
down vote
favorite
Find the range of the following function
$$f(x)=(x-7)(x+9)over(x-7)(x+1)$$
I tried to factor the two expressions(the way it is presented) in the numerator and denominator.
I dont know to proceed further.
Thanks for any help.
functions
asked 52 mins ago
Love Invariants
93215
2
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
1
I didnt see it. Thx..
– Love Invariants
44 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Find the range of the following function
$$f(x)=(x-7)(x+9)over(x-7)(x+1)$$
I tried to factor the two expressions(the way it is presented) in the numerator and denominator.
I dont know to proceed further.
Thanks for any help.
functions
asked 52 mins ago
Love Invariants
93215
Find the range of the following function
$$f(x)=(x-7)(x+9)over(x-7)(x+1)$$
I tried to factor the two expressions(the way it is presented) in the numerator and denominator.
I dont know to proceed further.
Thanks for any help.
functions
functions
asked 52 mins ago
Love Invariants
93215
asked 52 mins ago
Love Invariants
93215
edited 45 mins ago
asked 52 mins ago
Love Invariants
93215
asked 52 mins ago
Love Invariants
93215
asked 52 mins ago
Love Invariants
93215
93215
2
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
1
I didnt see it. Thx..
– Love Invariants
44 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago
add a comment |Â
2
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
1
I didnt see it. Thx..
– Love Invariants
44 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago
2
2
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
1
1
I didnt see it. Thx..
– Love Invariants
44 mins ago
I didnt see it. Thx..
– Love Invariants
44 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
The range is all real numbers except $y=1$ And $y=2$
Note that $y=1$ is horizontal asymptote and $y=2$ is a hole in the graph at $x=7$
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
add a comment |Â
up vote
2
down vote
Observing $fracx+9x+1=1+frac8x+1$ you will get that, as, $xin(-infty,infty)$, and easily you can see for $x=-1$ we will have an asymptote. Also, for $x=7$ we will have a hole in the graph. Hence, the range is $mathbbRbackslash2,1$.
answered 31 mins ago
tarit goswami
1,275219
add a comment |Â
up vote
0
down vote
The way to answer these types of questions you have to do the following provided $f$ is continuous on its domain:
- Determine the domain of the function
- Find the boundary of the domain
- Examine what happens near the boundary (consider left-hand and right-hand limits)
- Examine what happens at $pminfty$
- Also worry about local minima and local maxima if applicable.
In this case. The points that we need to examine are $7$ and $-1$. It appears that the function can be continuously extended at $x=7$, so this one doesn't give extra information. It is easy to see that
$$ lim_xuparrow-1 f(x) = -infty $$
and
$$ lim_xdownarrow-1 f(x) = +infty. $$
Let's see what happens now at the tails of the function.
$$ lim_xtopminfty f(x) = 1 $$.
Since the function doesn't have local maxima/minima (the function is everywhere monotone).
So from this we conclude that the range of $f$ on $(-infty,-1)$ is given by $(-infty,1)$ and on $(-1,infty)$ it is given by $(1,infty)$.
Concluding, the range of $f$ is given by $mathbbRsetminus1$.
Edit: Because of the comment I noticed I made a mistake. Why did I make the mistake? Because I just viewed the function as the continuous extension of $f$ which is something you usually do in practice. But to be precise, he/she is right and to solve this issue one finds the value of the continuous extension of $f$ at $7$ and this equals $2$. Since we concluded that $f$ was 'everywhere' monotone this just means that also $2$ is not part of its range. Hence the range is actually given by $mathbbRsetminus1,2$.
answered 36 mins ago
Stan Tendijck
1,336110
1
It appears that the function can be continuously extended at x=7- it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.
– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
add a comment |Â
up vote
0
down vote
For $xne7$, we have
beginalign
f(x) &= 1 + frac81+x\
x&in(-infty, infty)setminus7\
1+x&in(-infty, infty)setminus7
endalign
Since $frac11+x$ exists whenever $xne 0$,
beginalign
1+x&in(-infty,0)cup(0, infty)setminus7\
frac11+x &in (-infty,0)cup(0, infty)setminus7\
endalign
Edit: I had missed that $x=7$ is not in the domain (Since $frac1x-7$ is not defined at $7$. Hence $y=2$ will not be in the range.
Hence the required range in $mathbb Rsetminus2,7$
answered 36 mins ago
sc_
12112
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The range is all real numbers except $y=1$ And $y=2$
Note that $y=1$ is horizontal asymptote and $y=2$ is a hole in the graph at $x=7$
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
add a comment |Â
up vote
3
down vote
The range is all real numbers except $y=1$ And $y=2$
Note that $y=1$ is horizontal asymptote and $y=2$ is a hole in the graph at $x=7$
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The range is all real numbers except $y=1$ And $y=2$
Note that $y=1$ is horizontal asymptote and $y=2$ is a hole in the graph at $x=7$
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
The range is all real numbers except $y=1$ And $y=2$
Note that $y=1$ is horizontal asymptote and $y=2$ is a hole in the graph at $x=7$
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
answered 33 mins ago
Mohammad Riazi-Kermani
32.8k41853
32.8k41853
add a comment |Â
add a comment |Â
up vote
2
down vote
Observing $fracx+9x+1=1+frac8x+1$ you will get that, as, $xin(-infty,infty)$, and easily you can see for $x=-1$ we will have an asymptote. Also, for $x=7$ we will have a hole in the graph. Hence, the range is $mathbbRbackslash2,1$.
answered 31 mins ago
tarit goswami
1,275219
add a comment |Â
up vote
2
down vote
Observing $fracx+9x+1=1+frac8x+1$ you will get that, as, $xin(-infty,infty)$, and easily you can see for $x=-1$ we will have an asymptote. Also, for $x=7$ we will have a hole in the graph. Hence, the range is $mathbbRbackslash2,1$.
answered 31 mins ago
tarit goswami
1,275219
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Observing $fracx+9x+1=1+frac8x+1$ you will get that, as, $xin(-infty,infty)$, and easily you can see for $x=-1$ we will have an asymptote. Also, for $x=7$ we will have a hole in the graph. Hence, the range is $mathbbRbackslash2,1$.
answered 31 mins ago
tarit goswami
1,275219
Observing $fracx+9x+1=1+frac8x+1$ you will get that, as, $xin(-infty,infty)$, and easily you can see for $x=-1$ we will have an asymptote. Also, for $x=7$ we will have a hole in the graph. Hence, the range is $mathbbRbackslash2,1$.
answered 31 mins ago
tarit goswami
1,275219
edited 11 mins ago
answered 31 mins ago
tarit goswami
1,275219
answered 31 mins ago
tarit goswami
1,275219
answered 31 mins ago
tarit goswami
1,275219
1,275219
add a comment |Â
add a comment |Â
up vote
0
down vote
The way to answer these types of questions you have to do the following provided $f$ is continuous on its domain:
- Determine the domain of the function
- Find the boundary of the domain
- Examine what happens near the boundary (consider left-hand and right-hand limits)
- Examine what happens at $pminfty$
- Also worry about local minima and local maxima if applicable.
In this case. The points that we need to examine are $7$ and $-1$. It appears that the function can be continuously extended at $x=7$, so this one doesn't give extra information. It is easy to see that
$$ lim_xuparrow-1 f(x) = -infty $$
and
$$ lim_xdownarrow-1 f(x) = +infty. $$
Let's see what happens now at the tails of the function.
$$ lim_xtopminfty f(x) = 1 $$.
Since the function doesn't have local maxima/minima (the function is everywhere monotone).
So from this we conclude that the range of $f$ on $(-infty,-1)$ is given by $(-infty,1)$ and on $(-1,infty)$ it is given by $(1,infty)$.
Concluding, the range of $f$ is given by $mathbbRsetminus1$.
Edit: Because of the comment I noticed I made a mistake. Why did I make the mistake? Because I just viewed the function as the continuous extension of $f$ which is something you usually do in practice. But to be precise, he/she is right and to solve this issue one finds the value of the continuous extension of $f$ at $7$ and this equals $2$. Since we concluded that $f$ was 'everywhere' monotone this just means that also $2$ is not part of its range. Hence the range is actually given by $mathbbRsetminus1,2$.
answered 36 mins ago
Stan Tendijck
1,336110
1
It appears that the function can be continuously extended at x=7- it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.
– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
add a comment |Â
up vote
0
down vote
The way to answer these types of questions you have to do the following provided $f$ is continuous on its domain:
- Determine the domain of the function
- Find the boundary of the domain
- Examine what happens near the boundary (consider left-hand and right-hand limits)
- Examine what happens at $pminfty$
- Also worry about local minima and local maxima if applicable.
In this case. The points that we need to examine are $7$ and $-1$. It appears that the function can be continuously extended at $x=7$, so this one doesn't give extra information. It is easy to see that
$$ lim_xuparrow-1 f(x) = -infty $$
and
$$ lim_xdownarrow-1 f(x) = +infty. $$
Let's see what happens now at the tails of the function.
$$ lim_xtopminfty f(x) = 1 $$.
Since the function doesn't have local maxima/minima (the function is everywhere monotone).
So from this we conclude that the range of $f$ on $(-infty,-1)$ is given by $(-infty,1)$ and on $(-1,infty)$ it is given by $(1,infty)$.
Concluding, the range of $f$ is given by $mathbbRsetminus1$.
Edit: Because of the comment I noticed I made a mistake. Why did I make the mistake? Because I just viewed the function as the continuous extension of $f$ which is something you usually do in practice. But to be precise, he/she is right and to solve this issue one finds the value of the continuous extension of $f$ at $7$ and this equals $2$. Since we concluded that $f$ was 'everywhere' monotone this just means that also $2$ is not part of its range. Hence the range is actually given by $mathbbRsetminus1,2$.
answered 36 mins ago
Stan Tendijck
1,336110
1
It appears that the function can be continuously extended at x=7- it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.
– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The way to answer these types of questions you have to do the following provided $f$ is continuous on its domain:
- Determine the domain of the function
- Find the boundary of the domain
- Examine what happens near the boundary (consider left-hand and right-hand limits)
- Examine what happens at $pminfty$
- Also worry about local minima and local maxima if applicable.
In this case. The points that we need to examine are $7$ and $-1$. It appears that the function can be continuously extended at $x=7$, so this one doesn't give extra information. It is easy to see that
$$ lim_xuparrow-1 f(x) = -infty $$
and
$$ lim_xdownarrow-1 f(x) = +infty. $$
Let's see what happens now at the tails of the function.
$$ lim_xtopminfty f(x) = 1 $$.
Since the function doesn't have local maxima/minima (the function is everywhere monotone).
So from this we conclude that the range of $f$ on $(-infty,-1)$ is given by $(-infty,1)$ and on $(-1,infty)$ it is given by $(1,infty)$.
Concluding, the range of $f$ is given by $mathbbRsetminus1$.
Edit: Because of the comment I noticed I made a mistake. Why did I make the mistake? Because I just viewed the function as the continuous extension of $f$ which is something you usually do in practice. But to be precise, he/she is right and to solve this issue one finds the value of the continuous extension of $f$ at $7$ and this equals $2$. Since we concluded that $f$ was 'everywhere' monotone this just means that also $2$ is not part of its range. Hence the range is actually given by $mathbbRsetminus1,2$.
answered 36 mins ago
Stan Tendijck
1,336110
The way to answer these types of questions you have to do the following provided $f$ is continuous on its domain:
- Determine the domain of the function
- Find the boundary of the domain
- Examine what happens near the boundary (consider left-hand and right-hand limits)
- Examine what happens at $pminfty$
- Also worry about local minima and local maxima if applicable.
In this case. The points that we need to examine are $7$ and $-1$. It appears that the function can be continuously extended at $x=7$, so this one doesn't give extra information. It is easy to see that
$$ lim_xuparrow-1 f(x) = -infty $$
and
$$ lim_xdownarrow-1 f(x) = +infty. $$
Let's see what happens now at the tails of the function.
$$ lim_xtopminfty f(x) = 1 $$.
Since the function doesn't have local maxima/minima (the function is everywhere monotone).
So from this we conclude that the range of $f$ on $(-infty,-1)$ is given by $(-infty,1)$ and on $(-1,infty)$ it is given by $(1,infty)$.
Concluding, the range of $f$ is given by $mathbbRsetminus1$.
Edit: Because of the comment I noticed I made a mistake. Why did I make the mistake? Because I just viewed the function as the continuous extension of $f$ which is something you usually do in practice. But to be precise, he/she is right and to solve this issue one finds the value of the continuous extension of $f$ at $7$ and this equals $2$. Since we concluded that $f$ was 'everywhere' monotone this just means that also $2$ is not part of its range. Hence the range is actually given by $mathbbRsetminus1,2$.
answered 36 mins ago
Stan Tendijck
1,336110
edited 15 mins ago
answered 36 mins ago
Stan Tendijck
1,336110
answered 36 mins ago
Stan Tendijck
1,336110
answered 36 mins ago
Stan Tendijck
1,336110
1,336110
1
It appears that the function can be continuously extended at x=7- it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.
– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
add a comment |Â
1
It appears that the function can be continuously extended at x=7- it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.
– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
1
1
It appears that the function can be continuously extended at x=7 - it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.– tarit goswami
29 mins ago
It appears that the function can be continuously extended at x=7 - it could be extended, but this function is not defined like a step function. So, we will have a hole at $x=7$.– tarit goswami
29 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
You are right! Thanks for pointing it out. I changed my answer accordingly.
– Stan Tendijck
14 mins ago
add a comment |Â
up vote
0
down vote
For $xne7$, we have
beginalign
f(x) &= 1 + frac81+x\
x&in(-infty, infty)setminus7\
1+x&in(-infty, infty)setminus7
endalign
Since $frac11+x$ exists whenever $xne 0$,
beginalign
1+x&in(-infty,0)cup(0, infty)setminus7\
frac11+x &in (-infty,0)cup(0, infty)setminus7\
endalign
Edit: I had missed that $x=7$ is not in the domain (Since $frac1x-7$ is not defined at $7$. Hence $y=2$ will not be in the range.
Hence the required range in $mathbb Rsetminus2,7$
answered 36 mins ago
sc_
12112
add a comment |Â
up vote
0
down vote
For $xne7$, we have
beginalign
f(x) &= 1 + frac81+x\
x&in(-infty, infty)setminus7\
1+x&in(-infty, infty)setminus7
endalign
Since $frac11+x$ exists whenever $xne 0$,
beginalign
1+x&in(-infty,0)cup(0, infty)setminus7\
frac11+x &in (-infty,0)cup(0, infty)setminus7\
endalign
Edit: I had missed that $x=7$ is not in the domain (Since $frac1x-7$ is not defined at $7$. Hence $y=2$ will not be in the range.
Hence the required range in $mathbb Rsetminus2,7$
answered 36 mins ago
sc_
12112
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For $xne7$, we have
beginalign
f(x) &= 1 + frac81+x\
x&in(-infty, infty)setminus7\
1+x&in(-infty, infty)setminus7
endalign
Since $frac11+x$ exists whenever $xne 0$,
beginalign
1+x&in(-infty,0)cup(0, infty)setminus7\
frac11+x &in (-infty,0)cup(0, infty)setminus7\
endalign
Edit: I had missed that $x=7$ is not in the domain (Since $frac1x-7$ is not defined at $7$. Hence $y=2$ will not be in the range.
Hence the required range in $mathbb Rsetminus2,7$
answered 36 mins ago
sc_
12112
For $xne7$, we have
beginalign
f(x) &= 1 + frac81+x\
x&in(-infty, infty)setminus7\
1+x&in(-infty, infty)setminus7
endalign
Since $frac11+x$ exists whenever $xne 0$,
beginalign
1+x&in(-infty,0)cup(0, infty)setminus7\
frac11+x &in (-infty,0)cup(0, infty)setminus7\
endalign
Edit: I had missed that $x=7$ is not in the domain (Since $frac1x-7$ is not defined at $7$. Hence $y=2$ will not be in the range.
Hence the required range in $mathbb Rsetminus2,7$
answered 36 mins ago
sc_
12112
edited 12 mins ago
answered 36 mins ago
sc_
12112
answered 36 mins ago
sc_
12112
answered 36 mins ago
sc_
12112
12112
add a comment |Â
add a comment |Â
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2
Isn't $f(x)=1+dfrac8x+1$?
– giannispapav
47 mins ago
1
I didnt see it. Thx..
– Love Invariants
44 mins ago
By the way what values does this function not give. I see it giving all values. And there is (x-7)
– Love Invariants
42 mins ago
Since it's a rational function, everything except possible holes and horizontal asymptotes
– DWade64
37 mins ago