Let $f(x)=\sqrt{x^2+3x+4}$ be a rational-valued function of the rational variable $x$. Find the domain and range.

Let $s=\sqrt{x^2+3x+4}$. Then, $$s^2=x^2+3x+4\Rightarrow x^2+3x+4-s^2=0.$$ Since the discriminant has to be the square of a rational number, we have $$3^2-4\cdot 1\cdot (4-s^2)=t^2,\quad\text{i.e.}\quad 4s^2-7=t^2$$ for some $t\in\mathbb Q$.

So, we want $s,t\in\mathbb Q$ where $s\gt 0$ such that $$(2s)^2-t^2=7.$$

Now let us consider a hyperbola $x^2-y^2=7$. We want every rational $x\gt 0$.

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First of all, $(x,y)=(-4,3)$ is on the hyperbola. So, for $u\in\mathbb Q$, 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 $-1\lt u\lt 1$.

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 $x\gt 0$ such that $x^2-y^2=7$ where $y\in\mathbb Q$ can be written as $$\left\{x\mid x=\frac{4u^2+6u+4}{1-u^2},-1\lt u\lt 1,u\in\mathbb Q\right\}.$$

It follows from $s=\frac x2$ that the range of $f$ is $$\left\{y\mid y=\frac{2u^2+3u+2}{1-u^2},-1\lt u\lt 1,u\in\mathbb Q\right\},$$ i.e. $$\left\{y\mid y=\frac{p^2-3p+4}{2p-3},\color{red}{p\gt\frac 32},p\in\mathbb Q\right\}.$$

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 $x^2+3x+4-y^2=0$, $$x=\frac{-3\pm\sqrt{9-4\left(4-\left(\frac{p^2-3p+4}{2p-3}\right)^2\right)}}{2}=\frac{-3\pm\frac{2p^2-6p+1}{2p-3}}{2}=\frac{p^2-6p+5}{2p-3},\frac{4-p^2}{2p-3}$$

So, the domain of $f$ is $$\left\{x\mid x=\frac{p^2-6p+5}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\}\cup\left\{x\mid x=\frac{4-p^2}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\}.$$

By the way, we have $$\left\{x\mid x=\frac{p^2-6p+5}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\}=\left\{x\mid x=\frac{4-p^2}{2p-3},p\color{red}{\lt}\frac 32,p\in\mathbb Q\right\}$$ because $\frac{\color{blue}{(3-p)}^2-6\color{blue}{(3-p)}+5}{2\color{blue}{(3-p)}-3}=\frac{4-p^2}{2p-3}$.

It follows from this that the domain of $f$ is $$\left\{x\mid x=\frac{p^2-6p+5}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\}\cup\left\{x\mid x=\frac{4-p^2}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\},$$i.e. $$\left\{x\mid x=\frac{4-p^2}{2p-3},p\lt\frac 32,p\in\mathbb Q\right\}\cup\left\{x\mid x=\frac{4-p^2}{2p-3},p\gt\frac 32,p\in\mathbb Q\right\},$$ i.e.$$\left\{x\mid x=\frac{4-p^2}{2p-3},p\not=\frac 32,p\in\mathbb Q\right\}.$$