Logic symbols $\forall$ and $\exists$ What is the difference between the statements
$$\exists K> 0\, \forall x \in \mathbb{R}\, : |f(x)|< K   $$
and 
$$\forall x \in \mathbb{R}\, \exists K> 0\, : |f(x)|< K   $$
I know what $\forall$ and $\exists$ mean but I just can't see a difference between the two statements? I know that say $f(x)=x^4$ has the first property I just don't really understand why. Any help would be gratefully appreciated.
 A: Let's write the two statements with some paratheses: 
$$\tag 1 \exists K > 0 \, \bigl(\forall x \in \mathbf R: |f(x)| < K\bigr) $$
and 
$$ \tag 2 \forall x \in \mathbf R \, \bigl(\exists K > 0 : |f(x)| < K \bigr) $$
Let's start with (2). So forall $x \in \mathbf R$, it has to be true that 
$$ \tag{$2_x$} \exists K > 0 : |f(x)| < K $$
with is (obviously) true: $|f(x)|$ is a real number, there are larger ones, say $K = |f(x)| + 1$. So for each $x$ we can give a $K$. 
Now, let's look at (1). Here we have to give one $K$, such that 
$$ \tag{$1_K$} \forall x \in \mathbf R : |f(x)| < K $$
Note that $K$ must not depend on $x$, it may depend on $f$, of course, but depending on $f$, (1) may be false. $(1_K)$ says that $f$ is bounded by $K$, if $f$ is unbounded, as $f(x) = x^4$ is (1) is false. 
A: "There is a human $A$ such that for every other human $B$, $A$ is a parent of $B$."
"For every human $B$, there is a human $A$ such that $A$ is a parent of $B$."
I think you'll agree that the first is false and the second (in the absence of the Axiom of Death) is true.
A: With the first, you get a bounded function. The second is a null property, since you can take $K=2|f(x)|$ for every $x\in\Bbb R$.

And $f(x)=x^4$ is not a bounded function.
