Inequality, only one solution from algebra I recently came along the following problem:
$$f(x) = {4-x^2 \over 4-\sqrt{x}}$$
Solve for:$$f(x) ≥ 1$$
My Attempt
Now I know that one of the restrictions on the domain is $x≥0$, thus one of the solutions is $0≤x≤1$. There is still one more solution according to the graph, which is $x>16$.
Graph
Where did I go wrong?
 A: As already mentioned in the comments, you must also account for the case where $4-  \sqrt{x} < 0$, which results in the inequality sign being flipped.
The way to do this in general is to first get $0$ on one side and simplify/factor on the other:
\begin{align}
  \frac{4-x^2}{4-\sqrt x} &\ge 1\\[0.3cm]
  \frac{4-x^2}{4-\sqrt x} -1 &\ge 0\\[0.3cm]
  \frac{4-x^2}{4-\sqrt x} - \frac{4-\sqrt x}{4-\sqrt x} &\ge 0\\[0.3cm]
  \frac{4-x^2 - (4 - \sqrt x)}{4 - \sqrt x} &\ge 0\\[0.3cm]
  \frac{4-x^2 - 4 + \sqrt x}{4 - \sqrt x} &\ge 0\\[0.3cm]
  \frac{\sqrt x - x^2}{4 - \sqrt x} & \ge 0\\[0.3cm]
  \frac{\sqrt x (1 - x^{3/2})}{4 - \sqrt x} & \ge 0
\end{align}
Therefore solving the original equality is equivalent to solving this inequality.  Now we find all the values of $x$ where the numerator or denominator of the LHS is zero.  So we have three equations:
\begin{align}
  \sqrt x &= 0\\
  1 - x^{3/2} &= 0\\
  4 - \sqrt{x} &= 0
\end{align}
The first one gives us $x = 0$.  The second gives us $x = 1$.  The third gives us $x = 16$.  Now we use these values of $x$ to split the real number line into intervals:
$$ (0, 1) \qquad (1, 16) \qquad (16, +\infty) $$
Note that the LHS of our "new" inequality above cannot change sign within each of these intervals.  So we need only pick one point from each interval to determine the sign of the LHS.  Also note that we don't have $(-\infty, 0)$ because that's outside of the domain.
In $(0,1)$, let's pick $x = 1/2$.  We'll plug this in to the LHS above and see what its sign is.  We don't care about the value, only the sign:
$$\frac{\sqrt{1/2}(1 - [1/2]^{3/2})}{4 - \sqrt{1/2}} > 0$$
Proceed similarly in $(1,16)$ and $(16,+\infty)$ to see that the LHS is negative in $(1,16)$ and positive in $(16,+\infty)$.
Finally, note that the LHS is equal to zero when $x = 0$ and when $x = 1$.  Therefore the final answer is $[0,1] \cup (16, +\infty)$.
A: It is to be preferred in solving such inequalities to bring all terms to one side of the expression and make the comparison of one side to zero:
$$f(x) = {4-x^2 \over 4-\sqrt{x}} \ \ge \ 1 \ \ \Rightarrow \ \ {4-x^2 \over 4-\sqrt{x}} \ - \ 1 \  \ge \ 0 \ $$
$$ \frac{(4-x^2) \ - \ (4-\sqrt{x})}{4-\sqrt{x}} \ = \ \frac{\sqrt{x} \ - \ x^2}{4-\sqrt{x}} \ = \ \frac{\sqrt{x} \ ( 1 \ - \ x^{3/2})}{4-\sqrt{x}} \  \ge \ 0 \  \ . $$
There are two solutions to the equation portion, $ \ x \ = \ 0 \ $ and $ \ x \ = \ 1 \ $ , which you already know about.  There is also a "vertical asymptote" at $ \ x \ = \ 16 \ $  . So we need to inspect the signs of these factors in the intervals $ \ 0 \ < \ x \ < \  1 \ $ , $ \ 1 \ < \ x \ < \ 16 \ $  and  $ \ x \ > \ 16 \ $ .  ($ \ x  \ < \ 0 \ $ is excluded from the domain of $ \ f (x) \ $ .)
In order to have this ratio be positive, an odd number of the factors must be positive, which is to say that either just one must be, and all three must be.  This can only happen in the intervals $ \ 0 \ < \ x \ < \ 1 \ $ and $ \ x \  > \ 16 \ $ .  Putting these together with the permitted endpoints gives us the interval you found, $ \ 0 \ \le \ x \ \le \ 1 \ $ , and the last of these, $ \ x \ > \ 16 \ $ .
$ \ \ $ 

EDIT (14 May):  Here's another way to look at the inequality, which shows why it can be difficult to extract solutions directly.  To save a little writing, we will call the function $ \ 4 \ - \ x^2 \ $ in the numerator $ \ N ( x ) \ $ (marked in blue in the graph) and the denominator function $ \ D( x ) \ = \ 4 \ - \ \sqrt{x} \ $ (marked in red).  [The orange zone is outside the domain of $ \ D ( x ) \ $ .]
In the interval $ \ ( \ 0 \ , \ 1 \ ) \ $ , we have  $ \ \frac{N}{D} \ > \ 1 \ $ with $ \ N \ > \ D \ > \ 0 \ $ . There are of course intersections at $ \ x \ = \ 0 \ $ and $ \ x \ = \ 1 \ \ ( N \ = \ D ) \ $ . Beyond that (for a while), we have $ \ N \ < \ D \ $ , so our inequality is not satisfied; there is the point $ \ x \ = \ 2 \ $ beyond which $ \ N \ < \ 0 \ $ , but that has no special significance here. 
However, where it does become important is at the point $ \ x \ = \ 16 \ $ (not in the domain of $ \ D \ $ ), beyond which $ \ D \ < \ 0 \ $ .  So in the interval $ \ ( \ 16 \ , \ + \infty \ ) \ $ , we now have $ \ N \ < \ D \ < \ 0 \ $ , which produces the ratio $ \ \frac{N}{D} \ > \ 1 \ $ once more.  So this interval is the second portion of the solution set.  It is this change of sign in the denominator function that is easy to overlook in simplistic techniques.
The difficulty created by the approach that you took initially is that there is no good way to preserve the information about the special situation for $ \ x \ > \ 16 \ $ :  the cancelation in $$ \ 4 \ - \ x^2 \ > \ 4 \ - \ \sqrt{x} \ \ \Rightarrow \ \  -x^2 \ > \  -\sqrt{x} \ $$ simply obliterates it.  So this is not a method we encourage anyone to use.
A: No function may be divided by zero, so that fact must be reflected when considering the domain of the function.
So what must be true is the denominator of your f(x) must be greater than or equal to zero.  Not less than zero to stay within the domain of real numbers.  -sqrt(x) must be more than or equal to zero by algebraic manipulation, so hopefully then you can see how we get the solution x must be more than or equal to 16.
