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Let $f(x)=sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Find the domain and...
$begingroup$
Let $f(x)=sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Then find the domain and range of the function.
I tried to solve this problem but could not reach the answer. Since $x,f(x)in Bbb{Q}$, I let $f(x)=pin Bbb{Q}$.
$$p=sqrt{x^2+3x+4}Rightarrow p^2=x^2+3x+4Rightarrow x^2+3x+4-p^2=0$$
Since $x$ is a rational number,so the discriminant of the quadratic equation is perfect square and $geq0$.
$$9-4(4-p^2)geq0Rightarrow 4p^2-7geq0$$
But I am not getting the answer which is
domain $f$:
$$left{ {x~|~x=frac{4-p^2}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
range $f$:
$$left{ {y~|~y=frac{p^2-3p+4}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
Please help me.
functions elementary-set-theory
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
$endgroup$
add a comment |
$begingroup$
Let $f(x)=sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Then find the domain and range of the function.
I tried to solve this problem but could not reach the answer. Since $x,f(x)in Bbb{Q}$, I let $f(x)=pin Bbb{Q}$.
$$p=sqrt{x^2+3x+4}Rightarrow p^2=x^2+3x+4Rightarrow x^2+3x+4-p^2=0$$
Since $x$ is a rational number,so the discriminant of the quadratic equation is perfect square and $geq0$.
$$9-4(4-p^2)geq0Rightarrow 4p^2-7geq0$$
But I am not getting the answer which is
domain $f$:
$$left{ {x~|~x=frac{4-p^2}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
range $f$:
$$left{ {y~|~y=frac{p^2-3p+4}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
Please help me.
functions elementary-set-theory
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
$endgroup$
$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41
add a comment |
$begingroup$
Let $f(x)=sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Then find the domain and range of the function.
I tried to solve this problem but could not reach the answer. Since $x,f(x)in Bbb{Q}$, I let $f(x)=pin Bbb{Q}$.
$$p=sqrt{x^2+3x+4}Rightarrow p^2=x^2+3x+4Rightarrow x^2+3x+4-p^2=0$$
Since $x$ is a rational number,so the discriminant of the quadratic equation is perfect square and $geq0$.
$$9-4(4-p^2)geq0Rightarrow 4p^2-7geq0$$
But I am not getting the answer which is
domain $f$:
$$left{ {x~|~x=frac{4-p^2}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
range $f$:
$$left{ {y~|~y=frac{p^2-3p+4}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
Please help me.
functions elementary-set-theory
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
$endgroup$
Let $f(x)=sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Then find the domain and range of the function.
I tried to solve this problem but could not reach the answer. Since $x,f(x)in Bbb{Q}$, I let $f(x)=pin Bbb{Q}$.
$$p=sqrt{x^2+3x+4}Rightarrow p^2=x^2+3x+4Rightarrow x^2+3x+4-p^2=0$$
Since $x$ is a rational number,so the discriminant of the quadratic equation is perfect square and $geq0$.
$$9-4(4-p^2)geq0Rightarrow 4p^2-7geq0$$
But I am not getting the answer which is
domain $f$:
$$left{ {x~|~x=frac{4-p^2}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
range $f$:
$$left{ {y~|~y=frac{p^2-3p+4}{2p-3},~pnefrac{3}{2}},~pin Bbb{Q}right}$$
Please help me.
functions elementary-set-theory
functions elementary-set-theory
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
edited Sep 5 '15 at 14:10
Peter Woolfitt
18.6k54479
18.6k54479
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
asked Sep 5 '15 at 13:49
BrahmaguptaBrahmagupta
1,480827
1,480827
$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41
add a comment |
$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41
$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41
$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $s=sqrt{x^2+3x+4}$. Then,
$$s^2=x^2+3x+4Rightarrow x^2+3x+4-s^2=0.$$
Since the discriminant has to be the square of a rational number, we have
$$3^2-4cdot 1cdot (4-s^2)=t^2,quadtext{i.e.}quad 4s^2-7=t^2$$
for some $tinmathbb Q$.
So, we want $s,tinmathbb Q$ where $sgt 0$ such that
$$(2s)^2-t^2=7.$$
Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $xgt 0$.
$qquadqquadqquad$
First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $uinmathbb Q$, by Vieta's formulas, the $x$ coordinate $alpha$ of the intersection point other than $(-4,3)$ of the hyperbola with a line $y-3=u(x-(-4))$ is also rational :
$$x^2-(ux+4u+3)^2=7$$$$(1-u^2)x^2+(-8u^2-6u)x-16u^2-24u-16=0$$
$$-4+alpha=-frac{-8u^2-6u}{1-u^2}Rightarrow alpha=frac{4u^2+6u+4}{1-u^2}$$
In order for this to be positive, we need to have $-1lt ult 1$.
On the other hand, if the intersection point of the hyperbola with the line passing through $(-4,3)$ is a rational point, then the slope of the line is also rational.
Thus, we know that every rational $xgt 0$ such that $x^2-y^2=7$ where $yinmathbb Q$ can be written as
$$left{xmid x=frac{4u^2+6u+4}{1-u^2},-1lt ult 1,uinmathbb Qright}.$$
It follows from $s=frac x2$ that the range of $f$ is
$$left{ymid y=frac{2u^2+3u+2}{1-u^2},-1lt ult 1,uinmathbb Qright},$$
i.e.
$$left{ymid y=frac{p^2-3p+4}{2p-3},color{red}{pgtfrac 32},pinmathbb Qright}.$$
Here, I set $u=frac{2-p}{p-1}=-1+frac{1}{p-1}$. Note here that $p$ has to be larger than $frac 32$ because $y$ has to be positive.
Then, from $x^2+3x+4-y^2=0$,
$$x=frac{-3pmsqrt{9-4left(4-left(frac{p^2-3p+4}{2p-3}right)^2right)}}{2}=frac{-3pmfrac{2p^2-6p+1}{2p-3}}{2}=frac{p^2-6p+5}{2p-3},frac{4-p^2}{2p-3}$$
So, the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright}.$$
By the way, we have
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}=left{xmid x=frac{4-p^2}{2p-3},pcolor{red}{lt}frac 32,pinmathbb Qright}$$
because $frac{color{blue}{(3-p)}^2-6color{blue}{(3-p)}+5}{2color{blue}{(3-p)}-3}=frac{4-p^2}{2p-3}$.
It follows from this that the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$i.e.
$$left{xmid x=frac{4-p^2}{2p-3},pltfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$
i.e.$$left{xmid x=frac{4-p^2}{2p-3},pnot=frac 32,pinmathbb Qright}.$$
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Assume that
$$sqrt{x^2+3x+4}=ytag{1}$$
with rational $x$, $y$. Since $x^2+3x+4>0$ for real $x$ we have $y>0$. The rational numbers
$$u:={2xover y},qquad v:={2over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16over y^2}=4 ,$$
whence are rational points on the ellipse ${cal E}: u^2+3uv+4v^2=4$. Conversely: Any rational point $(u,v)in{cal E}$ with $v>0$ determines a solution of $(1)$ via $$x:={uover v},quad >y:={2over v} .tag{2}$$
In order to find the rational points on ${cal E}$ we note that $P:=(0,1)$ is such a point. If $W:=(u,v)$ is another rational point on ${cal E}$ then the line $Pvee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$ through $Pin{cal E}$ a second time with ${cal E}$ and obtain the point
$$(u,v)=left(-{3+8mover 1+3m+4m^2}, >{1-4m^2over 1+3m+4m^2}right)in{cal E}cap{mathbb Q}^2 ,$$
and all rational points on ${cal E}$ apart from $(0,-1)$ are obtained in this way. As we want $v>0$ we have to restrict $m$ to $-{1over2}<m<{1over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8mover 1-4m^2} ,quad y={2(1+3m+4m^2)over 1-4m^2}qquadleft(min bigl]-{1over2},{1over2}bigr[ cap{mathbb Q}right) .tag{3}$$
The domain of the function $f$ in question is then the set of $x$ generated by $(3)$, and the range of $f$ is the set of $y$ generated by $(3)$. Looking at the corresponding graphs one sees that the domain is unbounded below and above, while the range is bounded below by ${sqrt{7}over2}$, and unbounded above.
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
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add a comment |
$begingroup$
As f(x) is a real valued function the domain and range both should be real.
$f(x)=sqrt{ (x^2 + 3x + 4) }\
x^2 + 3x + 4 ≥ 0 $for f to be real
If we graph this function it is always positive as it's lowest point$ (frac{-3}{2},frac{7}{4}) $is positive.
The lowest point is calculated by differntiating the function and setting it equal to zero. It is the point where the function turns upwards.
Therefore in$ f(x) xepsilon R $
For range of $f(x)$
Let $\f(x)=sqrt{g(x)} where g(x)=x^2 + 3x + 4\
$
.Minimum value of $f(x)$ will be (minimum value of g(x))$^{({frac{1}{2}})}$
$\f(x)≥sqrt{frac{7}{4}}\
=> f(x)≥frac{sqrt7}{2}\$
Maximum value of f(x)=(maximum value of g(x))$^{frac{1}{2}}\
f(x)≤sqrt∞\
f(x)<∞. \\\$
$f(x)epsilon[frac{sqrt7}{2},∞)$ where $xepsilon R$
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
$endgroup$
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $s=sqrt{x^2+3x+4}$. Then,
$$s^2=x^2+3x+4Rightarrow x^2+3x+4-s^2=0.$$
Since the discriminant has to be the square of a rational number, we have
$$3^2-4cdot 1cdot (4-s^2)=t^2,quadtext{i.e.}quad 4s^2-7=t^2$$
for some $tinmathbb Q$.
So, we want $s,tinmathbb Q$ where $sgt 0$ such that
$$(2s)^2-t^2=7.$$
Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $xgt 0$.
$qquadqquadqquad$
First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $uinmathbb Q$, by Vieta's formulas, the $x$ coordinate $alpha$ of the intersection point other than $(-4,3)$ of the hyperbola with a line $y-3=u(x-(-4))$ is also rational :
$$x^2-(ux+4u+3)^2=7$$$$(1-u^2)x^2+(-8u^2-6u)x-16u^2-24u-16=0$$
$$-4+alpha=-frac{-8u^2-6u}{1-u^2}Rightarrow alpha=frac{4u^2+6u+4}{1-u^2}$$
In order for this to be positive, we need to have $-1lt ult 1$.
On the other hand, if the intersection point of the hyperbola with the line passing through $(-4,3)$ is a rational point, then the slope of the line is also rational.
Thus, we know that every rational $xgt 0$ such that $x^2-y^2=7$ where $yinmathbb Q$ can be written as
$$left{xmid x=frac{4u^2+6u+4}{1-u^2},-1lt ult 1,uinmathbb Qright}.$$
It follows from $s=frac x2$ that the range of $f$ is
$$left{ymid y=frac{2u^2+3u+2}{1-u^2},-1lt ult 1,uinmathbb Qright},$$
i.e.
$$left{ymid y=frac{p^2-3p+4}{2p-3},color{red}{pgtfrac 32},pinmathbb Qright}.$$
Here, I set $u=frac{2-p}{p-1}=-1+frac{1}{p-1}$. Note here that $p$ has to be larger than $frac 32$ because $y$ has to be positive.
Then, from $x^2+3x+4-y^2=0$,
$$x=frac{-3pmsqrt{9-4left(4-left(frac{p^2-3p+4}{2p-3}right)^2right)}}{2}=frac{-3pmfrac{2p^2-6p+1}{2p-3}}{2}=frac{p^2-6p+5}{2p-3},frac{4-p^2}{2p-3}$$
So, the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright}.$$
By the way, we have
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}=left{xmid x=frac{4-p^2}{2p-3},pcolor{red}{lt}frac 32,pinmathbb Qright}$$
because $frac{color{blue}{(3-p)}^2-6color{blue}{(3-p)}+5}{2color{blue}{(3-p)}-3}=frac{4-p^2}{2p-3}$.
It follows from this that the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$i.e.
$$left{xmid x=frac{4-p^2}{2p-3},pltfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$
i.e.$$left{xmid x=frac{4-p^2}{2p-3},pnot=frac 32,pinmathbb Qright}.$$
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Let $s=sqrt{x^2+3x+4}$. Then,
$$s^2=x^2+3x+4Rightarrow x^2+3x+4-s^2=0.$$
Since the discriminant has to be the square of a rational number, we have
$$3^2-4cdot 1cdot (4-s^2)=t^2,quadtext{i.e.}quad 4s^2-7=t^2$$
for some $tinmathbb Q$.
So, we want $s,tinmathbb Q$ where $sgt 0$ such that
$$(2s)^2-t^2=7.$$
Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $xgt 0$.
$qquadqquadqquad$
First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $uinmathbb Q$, by Vieta's formulas, the $x$ coordinate $alpha$ of the intersection point other than $(-4,3)$ of the hyperbola with a line $y-3=u(x-(-4))$ is also rational :
$$x^2-(ux+4u+3)^2=7$$$$(1-u^2)x^2+(-8u^2-6u)x-16u^2-24u-16=0$$
$$-4+alpha=-frac{-8u^2-6u}{1-u^2}Rightarrow alpha=frac{4u^2+6u+4}{1-u^2}$$
In order for this to be positive, we need to have $-1lt ult 1$.
On the other hand, if the intersection point of the hyperbola with the line passing through $(-4,3)$ is a rational point, then the slope of the line is also rational.
Thus, we know that every rational $xgt 0$ such that $x^2-y^2=7$ where $yinmathbb Q$ can be written as
$$left{xmid x=frac{4u^2+6u+4}{1-u^2},-1lt ult 1,uinmathbb Qright}.$$
It follows from $s=frac x2$ that the range of $f$ is
$$left{ymid y=frac{2u^2+3u+2}{1-u^2},-1lt ult 1,uinmathbb Qright},$$
i.e.
$$left{ymid y=frac{p^2-3p+4}{2p-3},color{red}{pgtfrac 32},pinmathbb Qright}.$$
Here, I set $u=frac{2-p}{p-1}=-1+frac{1}{p-1}$. Note here that $p$ has to be larger than $frac 32$ because $y$ has to be positive.
Then, from $x^2+3x+4-y^2=0$,
$$x=frac{-3pmsqrt{9-4left(4-left(frac{p^2-3p+4}{2p-3}right)^2right)}}{2}=frac{-3pmfrac{2p^2-6p+1}{2p-3}}{2}=frac{p^2-6p+5}{2p-3},frac{4-p^2}{2p-3}$$
So, the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright}.$$
By the way, we have
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}=left{xmid x=frac{4-p^2}{2p-3},pcolor{red}{lt}frac 32,pinmathbb Qright}$$
because $frac{color{blue}{(3-p)}^2-6color{blue}{(3-p)}+5}{2color{blue}{(3-p)}-3}=frac{4-p^2}{2p-3}$.
It follows from this that the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$i.e.
$$left{xmid x=frac{4-p^2}{2p-3},pltfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$
i.e.$$left{xmid x=frac{4-p^2}{2p-3},pnot=frac 32,pinmathbb Qright}.$$
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
$endgroup$
add a comment |
$begingroup$
Let $s=sqrt{x^2+3x+4}$. Then,
$$s^2=x^2+3x+4Rightarrow x^2+3x+4-s^2=0.$$
Since the discriminant has to be the square of a rational number, we have
$$3^2-4cdot 1cdot (4-s^2)=t^2,quadtext{i.e.}quad 4s^2-7=t^2$$
for some $tinmathbb Q$.
So, we want $s,tinmathbb Q$ where $sgt 0$ such that
$$(2s)^2-t^2=7.$$
Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $xgt 0$.
$qquadqquadqquad$
First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $uinmathbb Q$, by Vieta's formulas, the $x$ coordinate $alpha$ of the intersection point other than $(-4,3)$ of the hyperbola with a line $y-3=u(x-(-4))$ is also rational :
$$x^2-(ux+4u+3)^2=7$$$$(1-u^2)x^2+(-8u^2-6u)x-16u^2-24u-16=0$$
$$-4+alpha=-frac{-8u^2-6u}{1-u^2}Rightarrow alpha=frac{4u^2+6u+4}{1-u^2}$$
In order for this to be positive, we need to have $-1lt ult 1$.
On the other hand, if the intersection point of the hyperbola with the line passing through $(-4,3)$ is a rational point, then the slope of the line is also rational.
Thus, we know that every rational $xgt 0$ such that $x^2-y^2=7$ where $yinmathbb Q$ can be written as
$$left{xmid x=frac{4u^2+6u+4}{1-u^2},-1lt ult 1,uinmathbb Qright}.$$
It follows from $s=frac x2$ that the range of $f$ is
$$left{ymid y=frac{2u^2+3u+2}{1-u^2},-1lt ult 1,uinmathbb Qright},$$
i.e.
$$left{ymid y=frac{p^2-3p+4}{2p-3},color{red}{pgtfrac 32},pinmathbb Qright}.$$
Here, I set $u=frac{2-p}{p-1}=-1+frac{1}{p-1}$. Note here that $p$ has to be larger than $frac 32$ because $y$ has to be positive.
Then, from $x^2+3x+4-y^2=0$,
$$x=frac{-3pmsqrt{9-4left(4-left(frac{p^2-3p+4}{2p-3}right)^2right)}}{2}=frac{-3pmfrac{2p^2-6p+1}{2p-3}}{2}=frac{p^2-6p+5}{2p-3},frac{4-p^2}{2p-3}$$
So, the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright}.$$
By the way, we have
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}=left{xmid x=frac{4-p^2}{2p-3},pcolor{red}{lt}frac 32,pinmathbb Qright}$$
because $frac{color{blue}{(3-p)}^2-6color{blue}{(3-p)}+5}{2color{blue}{(3-p)}-3}=frac{4-p^2}{2p-3}$.
It follows from this that the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$i.e.
$$left{xmid x=frac{4-p^2}{2p-3},pltfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$
i.e.$$left{xmid x=frac{4-p^2}{2p-3},pnot=frac 32,pinmathbb Qright}.$$
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
$endgroup$
Let $s=sqrt{x^2+3x+4}$. Then,
$$s^2=x^2+3x+4Rightarrow x^2+3x+4-s^2=0.$$
Since the discriminant has to be the square of a rational number, we have
$$3^2-4cdot 1cdot (4-s^2)=t^2,quadtext{i.e.}quad 4s^2-7=t^2$$
for some $tinmathbb Q$.
So, we want $s,tinmathbb Q$ where $sgt 0$ such that
$$(2s)^2-t^2=7.$$
Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $xgt 0$.
$qquadqquadqquad$
First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $uinmathbb Q$, by Vieta's formulas, the $x$ coordinate $alpha$ of the intersection point other than $(-4,3)$ of the hyperbola with a line $y-3=u(x-(-4))$ is also rational :
$$x^2-(ux+4u+3)^2=7$$$$(1-u^2)x^2+(-8u^2-6u)x-16u^2-24u-16=0$$
$$-4+alpha=-frac{-8u^2-6u}{1-u^2}Rightarrow alpha=frac{4u^2+6u+4}{1-u^2}$$
In order for this to be positive, we need to have $-1lt ult 1$.
On the other hand, if the intersection point of the hyperbola with the line passing through $(-4,3)$ is a rational point, then the slope of the line is also rational.
Thus, we know that every rational $xgt 0$ such that $x^2-y^2=7$ where $yinmathbb Q$ can be written as
$$left{xmid x=frac{4u^2+6u+4}{1-u^2},-1lt ult 1,uinmathbb Qright}.$$
It follows from $s=frac x2$ that the range of $f$ is
$$left{ymid y=frac{2u^2+3u+2}{1-u^2},-1lt ult 1,uinmathbb Qright},$$
i.e.
$$left{ymid y=frac{p^2-3p+4}{2p-3},color{red}{pgtfrac 32},pinmathbb Qright}.$$
Here, I set $u=frac{2-p}{p-1}=-1+frac{1}{p-1}$. Note here that $p$ has to be larger than $frac 32$ because $y$ has to be positive.
Then, from $x^2+3x+4-y^2=0$,
$$x=frac{-3pmsqrt{9-4left(4-left(frac{p^2-3p+4}{2p-3}right)^2right)}}{2}=frac{-3pmfrac{2p^2-6p+1}{2p-3}}{2}=frac{p^2-6p+5}{2p-3},frac{4-p^2}{2p-3}$$
So, the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright}.$$
By the way, we have
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}=left{xmid x=frac{4-p^2}{2p-3},pcolor{red}{lt}frac 32,pinmathbb Qright}$$
because $frac{color{blue}{(3-p)}^2-6color{blue}{(3-p)}+5}{2color{blue}{(3-p)}-3}=frac{4-p^2}{2p-3}$.
It follows from this that the domain of $f$ is
$$left{xmid x=frac{p^2-6p+5}{2p-3},pgtfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$i.e.
$$left{xmid x=frac{4-p^2}{2p-3},pltfrac 32,pinmathbb Qright}cupleft{xmid x=frac{4-p^2}{2p-3},pgtfrac 32,pinmathbb Qright},$$
i.e.$$left{xmid x=frac{4-p^2}{2p-3},pnot=frac 32,pinmathbb Qright}.$$
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
edited Sep 17 '15 at 10:55
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
answered Sep 16 '15 at 10:09
mathlovemathlove
91.7k881215
91.7k881215
add a comment |
add a comment |
$begingroup$
Assume that
$$sqrt{x^2+3x+4}=ytag{1}$$
with rational $x$, $y$. Since $x^2+3x+4>0$ for real $x$ we have $y>0$. The rational numbers
$$u:={2xover y},qquad v:={2over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16over y^2}=4 ,$$
whence are rational points on the ellipse ${cal E}: u^2+3uv+4v^2=4$. Conversely: Any rational point $(u,v)in{cal E}$ with $v>0$ determines a solution of $(1)$ via $$x:={uover v},quad >y:={2over v} .tag{2}$$
In order to find the rational points on ${cal E}$ we note that $P:=(0,1)$ is such a point. If $W:=(u,v)$ is another rational point on ${cal E}$ then the line $Pvee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$ through $Pin{cal E}$ a second time with ${cal E}$ and obtain the point
$$(u,v)=left(-{3+8mover 1+3m+4m^2}, >{1-4m^2over 1+3m+4m^2}right)in{cal E}cap{mathbb Q}^2 ,$$
and all rational points on ${cal E}$ apart from $(0,-1)$ are obtained in this way. As we want $v>0$ we have to restrict $m$ to $-{1over2}<m<{1over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8mover 1-4m^2} ,quad y={2(1+3m+4m^2)over 1-4m^2}qquadleft(min bigl]-{1over2},{1over2}bigr[ cap{mathbb Q}right) .tag{3}$$
The domain of the function $f$ in question is then the set of $x$ generated by $(3)$, and the range of $f$ is the set of $y$ generated by $(3)$. Looking at the corresponding graphs one sees that the domain is unbounded below and above, while the range is bounded below by ${sqrt{7}over2}$, and unbounded above.
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Assume that
$$sqrt{x^2+3x+4}=ytag{1}$$
with rational $x$, $y$. Since $x^2+3x+4>0$ for real $x$ we have $y>0$. The rational numbers
$$u:={2xover y},qquad v:={2over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16over y^2}=4 ,$$
whence are rational points on the ellipse ${cal E}: u^2+3uv+4v^2=4$. Conversely: Any rational point $(u,v)in{cal E}$ with $v>0$ determines a solution of $(1)$ via $$x:={uover v},quad >y:={2over v} .tag{2}$$
In order to find the rational points on ${cal E}$ we note that $P:=(0,1)$ is such a point. If $W:=(u,v)$ is another rational point on ${cal E}$ then the line $Pvee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$ through $Pin{cal E}$ a second time with ${cal E}$ and obtain the point
$$(u,v)=left(-{3+8mover 1+3m+4m^2}, >{1-4m^2over 1+3m+4m^2}right)in{cal E}cap{mathbb Q}^2 ,$$
and all rational points on ${cal E}$ apart from $(0,-1)$ are obtained in this way. As we want $v>0$ we have to restrict $m$ to $-{1over2}<m<{1over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8mover 1-4m^2} ,quad y={2(1+3m+4m^2)over 1-4m^2}qquadleft(min bigl]-{1over2},{1over2}bigr[ cap{mathbb Q}right) .tag{3}$$
The domain of the function $f$ in question is then the set of $x$ generated by $(3)$, and the range of $f$ is the set of $y$ generated by $(3)$. Looking at the corresponding graphs one sees that the domain is unbounded below and above, while the range is bounded below by ${sqrt{7}over2}$, and unbounded above.
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Assume that
$$sqrt{x^2+3x+4}=ytag{1}$$
with rational $x$, $y$. Since $x^2+3x+4>0$ for real $x$ we have $y>0$. The rational numbers
$$u:={2xover y},qquad v:={2over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16over y^2}=4 ,$$
whence are rational points on the ellipse ${cal E}: u^2+3uv+4v^2=4$. Conversely: Any rational point $(u,v)in{cal E}$ with $v>0$ determines a solution of $(1)$ via $$x:={uover v},quad >y:={2over v} .tag{2}$$
In order to find the rational points on ${cal E}$ we note that $P:=(0,1)$ is such a point. If $W:=(u,v)$ is another rational point on ${cal E}$ then the line $Pvee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$ through $Pin{cal E}$ a second time with ${cal E}$ and obtain the point
$$(u,v)=left(-{3+8mover 1+3m+4m^2}, >{1-4m^2over 1+3m+4m^2}right)in{cal E}cap{mathbb Q}^2 ,$$
and all rational points on ${cal E}$ apart from $(0,-1)$ are obtained in this way. As we want $v>0$ we have to restrict $m$ to $-{1over2}<m<{1over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8mover 1-4m^2} ,quad y={2(1+3m+4m^2)over 1-4m^2}qquadleft(min bigl]-{1over2},{1over2}bigr[ cap{mathbb Q}right) .tag{3}$$
The domain of the function $f$ in question is then the set of $x$ generated by $(3)$, and the range of $f$ is the set of $y$ generated by $(3)$. Looking at the corresponding graphs one sees that the domain is unbounded below and above, while the range is bounded below by ${sqrt{7}over2}$, and unbounded above.
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
$endgroup$
Assume that
$$sqrt{x^2+3x+4}=ytag{1}$$
with rational $x$, $y$. Since $x^2+3x+4>0$ for real $x$ we have $y>0$. The rational numbers
$$u:={2xover y},qquad v:={2over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16over y^2}=4 ,$$
whence are rational points on the ellipse ${cal E}: u^2+3uv+4v^2=4$. Conversely: Any rational point $(u,v)in{cal E}$ with $v>0$ determines a solution of $(1)$ via $$x:={uover v},quad >y:={2over v} .tag{2}$$
In order to find the rational points on ${cal E}$ we note that $P:=(0,1)$ is such a point. If $W:=(u,v)$ is another rational point on ${cal E}$ then the line $Pvee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$ through $Pin{cal E}$ a second time with ${cal E}$ and obtain the point
$$(u,v)=left(-{3+8mover 1+3m+4m^2}, >{1-4m^2over 1+3m+4m^2}right)in{cal E}cap{mathbb Q}^2 ,$$
and all rational points on ${cal E}$ apart from $(0,-1)$ are obtained in this way. As we want $v>0$ we have to restrict $m$ to $-{1over2}<m<{1over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8mover 1-4m^2} ,quad y={2(1+3m+4m^2)over 1-4m^2}qquadleft(min bigl]-{1over2},{1over2}bigr[ cap{mathbb Q}right) .tag{3}$$
The domain of the function $f$ in question is then the set of $x$ generated by $(3)$, and the range of $f$ is the set of $y$ generated by $(3)$. Looking at the corresponding graphs one sees that the domain is unbounded below and above, while the range is bounded below by ${sqrt{7}over2}$, and unbounded above.
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
answered Sep 18 '15 at 14:10
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
$begingroup$
As f(x) is a real valued function the domain and range both should be real.
$f(x)=sqrt{ (x^2 + 3x + 4) }\
x^2 + 3x + 4 ≥ 0 $for f to be real
If we graph this function it is always positive as it's lowest point$ (frac{-3}{2},frac{7}{4}) $is positive.
The lowest point is calculated by differntiating the function and setting it equal to zero. It is the point where the function turns upwards.
Therefore in$ f(x) xepsilon R $
For range of $f(x)$
Let $\f(x)=sqrt{g(x)} where g(x)=x^2 + 3x + 4\
$
.Minimum value of $f(x)$ will be (minimum value of g(x))$^{({frac{1}{2}})}$
$\f(x)≥sqrt{frac{7}{4}}\
=> f(x)≥frac{sqrt7}{2}\$
Maximum value of f(x)=(maximum value of g(x))$^{frac{1}{2}}\
f(x)≤sqrt∞\
f(x)<∞. \\\$
$f(x)epsilon[frac{sqrt7}{2},∞)$ where $xepsilon R$
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
$endgroup$
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
add a comment |
$begingroup$
As f(x) is a real valued function the domain and range both should be real.
$f(x)=sqrt{ (x^2 + 3x + 4) }\
x^2 + 3x + 4 ≥ 0 $for f to be real
If we graph this function it is always positive as it's lowest point$ (frac{-3}{2},frac{7}{4}) $is positive.
The lowest point is calculated by differntiating the function and setting it equal to zero. It is the point where the function turns upwards.
Therefore in$ f(x) xepsilon R $
For range of $f(x)$
Let $\f(x)=sqrt{g(x)} where g(x)=x^2 + 3x + 4\
$
.Minimum value of $f(x)$ will be (minimum value of g(x))$^{({frac{1}{2}})}$
$\f(x)≥sqrt{frac{7}{4}}\
=> f(x)≥frac{sqrt7}{2}\$
Maximum value of f(x)=(maximum value of g(x))$^{frac{1}{2}}\
f(x)≤sqrt∞\
f(x)<∞. \\\$
$f(x)epsilon[frac{sqrt7}{2},∞)$ where $xepsilon R$
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
$endgroup$
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
add a comment |
$begingroup$
As f(x) is a real valued function the domain and range both should be real.
$f(x)=sqrt{ (x^2 + 3x + 4) }\
x^2 + 3x + 4 ≥ 0 $for f to be real
If we graph this function it is always positive as it's lowest point$ (frac{-3}{2},frac{7}{4}) $is positive.
The lowest point is calculated by differntiating the function and setting it equal to zero. It is the point where the function turns upwards.
Therefore in$ f(x) xepsilon R $
For range of $f(x)$
Let $\f(x)=sqrt{g(x)} where g(x)=x^2 + 3x + 4\
$
.Minimum value of $f(x)$ will be (minimum value of g(x))$^{({frac{1}{2}})}$
$\f(x)≥sqrt{frac{7}{4}}\
=> f(x)≥frac{sqrt7}{2}\$
Maximum value of f(x)=(maximum value of g(x))$^{frac{1}{2}}\
f(x)≤sqrt∞\
f(x)<∞. \\\$
$f(x)epsilon[frac{sqrt7}{2},∞)$ where $xepsilon R$
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
$endgroup$
As f(x) is a real valued function the domain and range both should be real.
$f(x)=sqrt{ (x^2 + 3x + 4) }\
x^2 + 3x + 4 ≥ 0 $for f to be real
If we graph this function it is always positive as it's lowest point$ (frac{-3}{2},frac{7}{4}) $is positive.
The lowest point is calculated by differntiating the function and setting it equal to zero. It is the point where the function turns upwards.
Therefore in$ f(x) xepsilon R $
For range of $f(x)$
Let $\f(x)=sqrt{g(x)} where g(x)=x^2 + 3x + 4\
$
.Minimum value of $f(x)$ will be (minimum value of g(x))$^{({frac{1}{2}})}$
$\f(x)≥sqrt{frac{7}{4}}\
=> f(x)≥frac{sqrt7}{2}\$
Maximum value of f(x)=(maximum value of g(x))$^{frac{1}{2}}\
f(x)≤sqrt∞\
f(x)<∞. \\\$
$f(x)epsilon[frac{sqrt7}{2},∞)$ where $xepsilon R$
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
answered Jan 8 at 5:56
Swap NayakSwap Nayak
22
22
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
add a comment |
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
$begingroup$
I did it for a real valued function
$endgroup$
– Swap Nayak
Jan 8 at 6:03
add a comment |
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$begingroup$
but i'm getting $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$ or $$x=frac{1}{2} left(-sqrt{4 p^2-7}-3right)$$
$endgroup$
– Dr. Sonnhard Graubner
Sep 5 '15 at 14:41