Does the first derivative test always work for finding minima and maxima? Suppose you want to find the max of the function $\ f(x)=\sqrt{x} - x$.
Using the first derivative test you get, $f'(x)= \frac{1}{2\sqrt{x}} - 1$ .
If we equate this to $0$ we get $x =\frac{1}{4}$. Taking as $x$ as $0.20$ and $0.30$, we get that the first derivative doesn't change signs (remains positive). However, if x is taken as $x > 1$, then the first derivative becomes negative.
Graphing the function reveals that $x =\frac{1}{4}$ is indeed the maximum point. 
Taking the second derivative $\ f\prime\prime(x)= \frac{-1}{4x^\frac{3}{2}}$ and using the second derivative test at the point $x =\frac{1}{4}$ shows the second derivative is negative, indicating a maximum.
My question therefore is does the first derivative test necessarily always show the maximum? Both the graphs and second derivatives indicate a maximum; however if the first derivative is taken with $x < 1$ then the first derivative test fails. Can someone explain how this could happen?

As I said in a comment, your calculations are wrong. $f'(0.2)$ is indeed positive, but $f'(0.3)$ is negative.
The first derivative test (checking how $f'$ changes sign at a critical point) and the second derivative test (checking the sign of $f''$ at a critical point) will never give contradictory conclusions, as you thought they did here. (You thought the first derivative was indicating a non-extremum whereas the second derivative was indicating a maximum. But your evidence was wrong, because $f'(x)$ does indeed change sign at the critical point.)
The second derivative test may be inconclusive (this happens precisely when $f''$ vanishes at a critical point), but you will never get contradictory results.
