Show that extension $\mathbb{R}(x) \supseteq \mathbb{R}$ is a real extension (ordered extension) 
Problem: Denote $\mathbb{R}(x)$ be the quotient field of polynomial ring in single variable $\mathbb{R}[x]$. Show that extension $\mathbb{R}(x) \supseteq \mathbb{R}$ is a real extension (ordered extension).

My attempt: Follows the definition of real extension, if we denote ordering on $\mathbb{R}(x)$ is $\le_{\mathbb{R}(x)}$ then $\mathbb{R}(x) \supseteq \mathbb{R}$ is a real extension iff the restriction of $\le_{\mathbb{R}(x)}$ on $\mathbb{R}$ is equal to the ordering on $\mathbb{R}$. Let $f(x), f'(x), g(x), g'(x) \in \mathbb{R}[x]$ and $\frac{f(x)}{g(x)} \le_{\mathbb{R}(x)} \frac{f'(x)}{g'(x)}$. Let $g(x) = 1, \forall x$ then we have $f(x) \le f'(x)$. Q.E.D
 A: So I don't believe there is a common enough ordering on $\mathbb{R}[x]$ to be called the "usual" one. For real-closed fields, such as $\mathbb{R}$ the ordering is uniquely defined by $a > 0$ if $a = b^2$ for some $b$.
For $\mathbb{R}(x)$ since $x$ is not a square or a sum of squares, there's nothing to say whether we should have $x > 0$ or $x < 0$. In fact, given any ordering of $\mathbb{R}(x)$ we can swap $x$ and $-x$ an get a different ordering in which the sign of $x$ is swapped.
I would say the easiest way to see that $\mathbb{R}(x)$ is formally-real isn't via any order but to see that $-1$ isn't a sum of squares. For suppose
$$ -1 = \sum \left(\frac{f_i(x)}{g_i(x)}\right)^2 $$
and pick $a \in \mathbb{R}$ so that $g_i(a) \ne 0$ for any $i$. Then
$$ -1 = \sum \left(\frac{f_i(a)}{g_i(a)}\right)^2, $$
which is impossible.
Wikipedia describes an ordering on $\mathbb{R}(x)$ where the positive cone is
$$\left\{\frac fg : \frac{\text{leading coefficient of }f}{\text{leading coefficient of }g} > 0 \right\}.$$
But like I said, this ordering isn't unique. One then has to check that this ordering makes $\mathbb{R}(x)$ into an ordered field. This ordering obviously extends the ordering on $\mathbb{R}$.
Although, you don't really need to think about whether the ordering extends that of $\mathbb{R}$. No matter what ordering you have, sums of squares are always positive and in $\mathbb{R}$, the positive cone is exactly the set of squares. Thus any ordering on some field extension of $\mathbb{R}$ will always extend the order on $\mathbb{R}$.
