Let $f(x)=\sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Find the domain and range. 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 $f(x)=p\in \Bbb{Q}$.
$$p=\sqrt{x^2+3x+4}\Rightarrow p^2=x^2+3x+4\Rightarrow x^2+3x+4-p^2=0$$
Since $x$ is a rational number,so the discriminant of the quadratic equation is perfect square and $\geq0$.
$$9-4(4-p^2)\geq0\Rightarrow 4p^2-7\geq0$$ 
But I am not getting the answer which is
domain $f$: 
$$\left\{ {x~|~x=\frac{4-p^2}{2p-3},~p\ne\frac{3}{2}},~p\in \Bbb{Q}\right\}$$
range $f$:
$$\left\{ {y~|~y=\frac{p^2-3p+4}{2p-3},~p\ne\frac{3}{2}},~p\in \Bbb{Q}\right\}$$
Please help me.
 A: Assume that
$$\sqrt{x^2+3x+4}=y\tag{1}$$
with rational $x$, $y$.  Since $x^2+3x+4>0$ for real $x$ we  have $y>0$. The rational numbers
$$u:={2x\over y},\qquad v:={2\over y}$$
then satisfy
$$u^2+3uv+4v^2={4x^2+12x+16\over y^2}=4\ ,$$
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 $$x:={u\over v},\quad \>y:={2\over v}\ .\tag{2}$$ 
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 $P\vee W$ has rational slope $m$. We therefore intersect the line $v= 1+m u$  through $P\in{\cal E}$ a second time with ${\cal E}$ and obtain the point
$$(u,v)=\left(-{3+8m\over 1+3m+4m^2}, \>{1-4m^2\over 1+3m+4m^2}\right)\in{\cal E}\cap{\mathbb Q}^2\ ,$$
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 $-{1\over2}<m<{1\over2}$. In this way we obtain from $(2)$ the following parametrization of the rational solutions of $(1)$:
$$x=-{3+8m\over 1-4m^2}\ ,\quad y={2(1+3m+4m^2)\over 1-4m^2}\qquad\left(m\in\ \bigl]-{1\over2},{1\over2}\bigr[\ \cap{\mathbb Q}\right)\ .\tag{3}$$
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.
A: 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) x\epsilon 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 $x\epsilon R$
