Verifying linear independence for $\left\{ \sqrt{x} , \sqrt{x+1}, \sqrt{x+2} \right\}$ using Wronskian I need to verify the linear independence for the group of functions:
 $$\left\{ \;f_1 = \sqrt{x} , \;f_2 = \sqrt{x+1}, \;f_3 = \sqrt{x+2} \right\}$$
using Wronskian, for $x > 0$.  

I wasn't told whether those functions are a solution of an ODE!
So I can't use the theorem which holds for a set of solutions, which says: $$W[f_1,f_2,f_3](x_0 \in I) \ne 0 \iff (f_1,f_2,f_3) \text{ form a basis}$$

The Wronskian is:
$$
W = \begin{vmatrix}
\sqrt{x} & \sqrt{x+1} & \sqrt{x+2} \\
\frac{1}{2\sqrt{x}} & \frac{1}{2\sqrt{x+1}} & \frac{1}{2\sqrt{x+2}} \\
\frac{-1}{4(\sqrt{x})^3} & \frac{-1}{4(\sqrt{x+1})^3} & \frac{-1}{4(\sqrt{x + 2})^3}
\end{vmatrix}
$$
So its final expression is really messy and isn't reduced to zero or anything "nice".
If I were to plug $x=1$, then the determinant would be non-zero, see here.  
I've seen here the fact that:  

Given two functions $f(x)$ and $g(x)$ are differentiable on some
  interval $I$.
  If $W(f,g)(x_0) \ne 0$ for some $x_0$ in $I$, then
  $f(x)$ and $g(x)$ are linearly independent on the interval $I$.

But in my notebook, the same sentence is applied to a set of functions which are a solution of a linear homogeneous ODE, which apply the theorem on existence and uniqueness.
While in this case, the set of functions isn't necessarily a solution of an ODE. I was told only to test for their linearly dependency and nothing extra was told about them.
So how would you assert for the linear independence of the given set of functions?
 A: Don't make your life more complicated than necessary: 

The Wronskian can test any finite family of (sufficiently differenetiable) functions for linear independence. 

Assume $f_1, \ldots, f_n$ are $n-1$ times differentiable and linearly dependent, say $f_n$ can be expressed as a linear combination $f_n=c_1f_1+\ldots +c_{n-1}f_{n-1}$, then the coerresponding relation holds for $f_n'=c_1f_1'+\ldots +c_{n-1}f_{n-1}'$ up to $f_n^{(n-1)}=c_1f_1^{(n-1)}+\ldots +c_{n-1}f_{n-1}^{(n-1)}$. Then the $n$th column of the Wronskian is a linear combination of the other columns, hence the determinant is zero, no matter what $x$ you plug in. Hence conversely, if the Wronskian is $\ne0$ for even a single  $x_0$, the functions are linearly independant.

On the other hand, a direct approach might be less cumbersome in this special case than the generic method using the Wronskian:
Assume $c_1\sqrt x+c_2\sqrt{x+1}+c_3\sqrt{x+2}=0$ for all $x>0$. Taking $\lim_{x\to0^+}$ we find $c_2+c_3\sqrt 2=0$.  Taking $\lim_{x\to 0^+}$ of the derivatives we  conclude $c_1=0$ because only the first function has unbounded derivative near $0$. Hence $c_3\cdot(\sqrt{x+2}-\sqrt2\sqrt{x+1})=0$. Plug in $x=1$ to see $c_3\cdot(\sqrt3-2)=0$,hence $c_3=0$ and also $c_2=-\sqrt 2c_3=0$.
