"Determine whether $f(x) = x^3, g(x) = x^2|x|$ are linearly independent on the real line."
My Work:
Independent if Wronskian $(f,g) ≠ 0$ for all points (Note: the square brackets are a determinant)
Wronskian, $$W(f,g) = \begin{bmatrix}f & g\\f' & g'\end{bmatrix} $$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\begin{bmatrix}x^3& x^2|x|\\3x^2 & 2x|x| + \frac{x^3}{|x|}\end{bmatrix} $$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2x^4|x| + \frac{x^6}{|x|}-3x^4|x|$$
Set equal to $0$, and can then divide by $x^4$ for simplicity: $$2|x| + \frac{x^2}{|x|}-3|x| = 0$$ We can see that if $x = 1$, equation is valid, therefore I expect this to NOT be linearly independent. But apparently, this IS linearly independent, although no explanation is provided. Where have I gone wrong?
You do not need the Wronskian to be nonzero at all points: it's sufficient for the Wronskian to be nonzero at one point. Said another way, if functions are linearly dependent then their Wronskian is identically zero. The contrapositive of this statement is that if the Wronskian is not identically zero, meaning it's nonzero somewhere, then the functions are linearly independent.
It's a very, very special property of functions which are solutions to the same (linear, constant-coefficient) differential equation that if their Wronskian is nonzero it's nonzero everywhere.
Incidentally, $\frac{x^2}{|x|} = |x|$, so the Wronskian actually vanishes identically in this case (you need to be a little careful about what happens at $x = 0$ but it checks out). So you actually cannot use the Wronskian here and have to try something else.