Proving that $fcolon mathbb{R} to mathbb{R}$ $frac{x}{1+x^2}$ is not an injection
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I needed help proving that this function is not an injection. I know that I can try a few $x$ values and see if I get the same output but I wanted a more rigorous proof using the definitions.
This is what I tried.
begin{align}
f(m)=f(p)
&implies frac{m}{1+m^2} = frac{p}{1+p^2} \
&implies m+mp^2 = p + pm^2 \
&implies m(1+p^2) = p(1+m^2)
end{align}
I don't really know where to go from here, how would I derive a contradiction showing that this is not injective?
functions proof-writing
add a comment |
up vote
1
down vote
favorite
I needed help proving that this function is not an injection. I know that I can try a few $x$ values and see if I get the same output but I wanted a more rigorous proof using the definitions.
This is what I tried.
begin{align}
f(m)=f(p)
&implies frac{m}{1+m^2} = frac{p}{1+p^2} \
&implies m+mp^2 = p + pm^2 \
&implies m(1+p^2) = p(1+m^2)
end{align}
I don't really know where to go from here, how would I derive a contradiction showing that this is not injective?
functions proof-writing
1
You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I needed help proving that this function is not an injection. I know that I can try a few $x$ values and see if I get the same output but I wanted a more rigorous proof using the definitions.
This is what I tried.
begin{align}
f(m)=f(p)
&implies frac{m}{1+m^2} = frac{p}{1+p^2} \
&implies m+mp^2 = p + pm^2 \
&implies m(1+p^2) = p(1+m^2)
end{align}
I don't really know where to go from here, how would I derive a contradiction showing that this is not injective?
functions proof-writing
I needed help proving that this function is not an injection. I know that I can try a few $x$ values and see if I get the same output but I wanted a more rigorous proof using the definitions.
This is what I tried.
begin{align}
f(m)=f(p)
&implies frac{m}{1+m^2} = frac{p}{1+p^2} \
&implies m+mp^2 = p + pm^2 \
&implies m(1+p^2) = p(1+m^2)
end{align}
I don't really know where to go from here, how would I derive a contradiction showing that this is not injective?
functions proof-writing
functions proof-writing
edited Nov 14 at 15:46
Xander Henderson
13.8k93552
13.8k93552
asked Nov 14 at 3:17
Rufyi
647
647
1
You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22
add a comment |
1
You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22
1
1
You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22
You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22
add a comment |
5 Answers
5
active
oldest
votes
up vote
7
down vote
accepted
Note that for $x ne 0$, we have $$frac{1/x}{1 + (1/x)^2} = frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = frac{x}{1+x^2} = frac{x/x}{1/x + x^2/x} = frac{1}{x + frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
add a comment |
up vote
5
down vote
Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly.
add a comment |
up vote
4
down vote
I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer. The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever. The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,yinmathbb{R}$ (where these two values are not assumed to be distinct). Then we have
$$ frac{x}{1+x^2} = frac{y}{1+y^2}
implies x(1+y^2) = y(1+x^2)
implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
begin{align} y
&= frac{(1+x^2) pm sqrt{(1+x^2)^2 - 4x^2}}{2x} \
&= frac{(1+x^2) pm sqrt{1-2x^2+x^4}}{2x} \
&= frac{(1+x^2) pm sqrt{(1-x^2)^2}}{2x} \
&= frac{(1+x^2) pm (1-x^2)}{2x}.
end{align}
Therefore, simplifying the $pm$, we conclude that
$$ y in left{ frac{(1+x^2) + (1-x^2)}{2x}, frac{(1+x^2) - (1-x^2)}{2x} right}
= left{ frac{1}{x}, xright}.
$$
As long as $x ne 0$, we obtain two possible values of $y$: either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $xne y$ (demonstrating a lack of injectivity). Indeed, what we have now shown is exactly heropup's solution. If we take $y=frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= frac{1/x}{1+(1/x)^2}
= frac{x}{1+x^2}
= f(x)
$$
for all $xne 0$.
add a comment |
up vote
2
down vote
Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
add a comment |
up vote
1
down vote
Note that $f(0)=0$ and $lim_{x to infty} f(x)=0$. There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +infty)$. Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons. (This is a little flabby, but can be made rigorous.)
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Note that for $x ne 0$, we have $$frac{1/x}{1 + (1/x)^2} = frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = frac{x}{1+x^2} = frac{x/x}{1/x + x^2/x} = frac{1}{x + frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
add a comment |
up vote
7
down vote
accepted
Note that for $x ne 0$, we have $$frac{1/x}{1 + (1/x)^2} = frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = frac{x}{1+x^2} = frac{x/x}{1/x + x^2/x} = frac{1}{x + frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Note that for $x ne 0$, we have $$frac{1/x}{1 + (1/x)^2} = frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = frac{x}{1+x^2} = frac{x/x}{1/x + x^2/x} = frac{1}{x + frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
Note that for $x ne 0$, we have $$frac{1/x}{1 + (1/x)^2} = frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = frac{x}{1+x^2} = frac{x/x}{1/x + x^2/x} = frac{1}{x + frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
edited Nov 14 at 3:26
answered Nov 14 at 3:24
heropup
61.9k65997
61.9k65997
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
add a comment |
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
1
1
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
This is amazingly slick. I wouldn't've thought of this in a hundred years.
– Randall
Nov 14 at 3:25
1
1
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall It is actually not surprising that you wouldn't have thought of it. Unless you have seen problems like this before, it is not at all obvious that this kind of symmetry would show up. Once you know the trick, it is kind of slick to use it, but it takes some work to get there.
– Xander Henderson
Nov 14 at 3:47
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
@Randall: See my answer for how to derive the answer.
– user21820
Nov 14 at 5:11
add a comment |
up vote
5
down vote
Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly.
add a comment |
up vote
5
down vote
Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly.
add a comment |
up vote
5
down vote
up vote
5
down vote
Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly.
Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly.
answered Nov 14 at 3:21
Zachary Selk
398211
398211
add a comment |
add a comment |
up vote
4
down vote
I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer. The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever. The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,yinmathbb{R}$ (where these two values are not assumed to be distinct). Then we have
$$ frac{x}{1+x^2} = frac{y}{1+y^2}
implies x(1+y^2) = y(1+x^2)
implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
begin{align} y
&= frac{(1+x^2) pm sqrt{(1+x^2)^2 - 4x^2}}{2x} \
&= frac{(1+x^2) pm sqrt{1-2x^2+x^4}}{2x} \
&= frac{(1+x^2) pm sqrt{(1-x^2)^2}}{2x} \
&= frac{(1+x^2) pm (1-x^2)}{2x}.
end{align}
Therefore, simplifying the $pm$, we conclude that
$$ y in left{ frac{(1+x^2) + (1-x^2)}{2x}, frac{(1+x^2) - (1-x^2)}{2x} right}
= left{ frac{1}{x}, xright}.
$$
As long as $x ne 0$, we obtain two possible values of $y$: either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $xne y$ (demonstrating a lack of injectivity). Indeed, what we have now shown is exactly heropup's solution. If we take $y=frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= frac{1/x}{1+(1/x)^2}
= frac{x}{1+x^2}
= f(x)
$$
for all $xne 0$.
add a comment |
up vote
4
down vote
I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer. The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever. The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,yinmathbb{R}$ (where these two values are not assumed to be distinct). Then we have
$$ frac{x}{1+x^2} = frac{y}{1+y^2}
implies x(1+y^2) = y(1+x^2)
implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
begin{align} y
&= frac{(1+x^2) pm sqrt{(1+x^2)^2 - 4x^2}}{2x} \
&= frac{(1+x^2) pm sqrt{1-2x^2+x^4}}{2x} \
&= frac{(1+x^2) pm sqrt{(1-x^2)^2}}{2x} \
&= frac{(1+x^2) pm (1-x^2)}{2x}.
end{align}
Therefore, simplifying the $pm$, we conclude that
$$ y in left{ frac{(1+x^2) + (1-x^2)}{2x}, frac{(1+x^2) - (1-x^2)}{2x} right}
= left{ frac{1}{x}, xright}.
$$
As long as $x ne 0$, we obtain two possible values of $y$: either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $xne y$ (demonstrating a lack of injectivity). Indeed, what we have now shown is exactly heropup's solution. If we take $y=frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= frac{1/x}{1+(1/x)^2}
= frac{x}{1+x^2}
= f(x)
$$
for all $xne 0$.
add a comment |
up vote
4
down vote
up vote
4
down vote
I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer. The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever. The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,yinmathbb{R}$ (where these two values are not assumed to be distinct). Then we have
$$ frac{x}{1+x^2} = frac{y}{1+y^2}
implies x(1+y^2) = y(1+x^2)
implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
begin{align} y
&= frac{(1+x^2) pm sqrt{(1+x^2)^2 - 4x^2}}{2x} \
&= frac{(1+x^2) pm sqrt{1-2x^2+x^4}}{2x} \
&= frac{(1+x^2) pm sqrt{(1-x^2)^2}}{2x} \
&= frac{(1+x^2) pm (1-x^2)}{2x}.
end{align}
Therefore, simplifying the $pm$, we conclude that
$$ y in left{ frac{(1+x^2) + (1-x^2)}{2x}, frac{(1+x^2) - (1-x^2)}{2x} right}
= left{ frac{1}{x}, xright}.
$$
As long as $x ne 0$, we obtain two possible values of $y$: either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $xne y$ (demonstrating a lack of injectivity). Indeed, what we have now shown is exactly heropup's solution. If we take $y=frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= frac{1/x}{1+(1/x)^2}
= frac{x}{1+x^2}
= f(x)
$$
for all $xne 0$.
I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer. The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever. The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,yinmathbb{R}$ (where these two values are not assumed to be distinct). Then we have
$$ frac{x}{1+x^2} = frac{y}{1+y^2}
implies x(1+y^2) = y(1+x^2)
implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
begin{align} y
&= frac{(1+x^2) pm sqrt{(1+x^2)^2 - 4x^2}}{2x} \
&= frac{(1+x^2) pm sqrt{1-2x^2+x^4}}{2x} \
&= frac{(1+x^2) pm sqrt{(1-x^2)^2}}{2x} \
&= frac{(1+x^2) pm (1-x^2)}{2x}.
end{align}
Therefore, simplifying the $pm$, we conclude that
$$ y in left{ frac{(1+x^2) + (1-x^2)}{2x}, frac{(1+x^2) - (1-x^2)}{2x} right}
= left{ frac{1}{x}, xright}.
$$
As long as $x ne 0$, we obtain two possible values of $y$: either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $xne y$ (demonstrating a lack of injectivity). Indeed, what we have now shown is exactly heropup's solution. If we take $y=frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= frac{1/x}{1+(1/x)^2}
= frac{x}{1+x^2}
= f(x)
$$
for all $xne 0$.
edited Nov 14 at 15:42
answered Nov 14 at 4:00
Xander Henderson
13.8k93552
13.8k93552
add a comment |
add a comment |
up vote
2
down vote
Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
add a comment |
up vote
2
down vote
Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
add a comment |
up vote
2
down vote
up vote
2
down vote
Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
answered Nov 14 at 5:09
user21820
38k441149
38k441149
add a comment |
add a comment |
up vote
1
down vote
Note that $f(0)=0$ and $lim_{x to infty} f(x)=0$. There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +infty)$. Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons. (This is a little flabby, but can be made rigorous.)
add a comment |
up vote
1
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Note that $f(0)=0$ and $lim_{x to infty} f(x)=0$. There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +infty)$. Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons. (This is a little flabby, but can be made rigorous.)
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Note that $f(0)=0$ and $lim_{x to infty} f(x)=0$. There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +infty)$. Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons. (This is a little flabby, but can be made rigorous.)
Note that $f(0)=0$ and $lim_{x to infty} f(x)=0$. There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +infty)$. Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons. (This is a little flabby, but can be made rigorous.)
answered Nov 14 at 3:24
Randall
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8,49411128
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You could solve the quadratic $m+mp^2 = p + pm^2$ for either $p$ or $m$, then choose the other variable (more or less) freely.
– Xander Henderson
Nov 14 at 3:22