<s>[INST] 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?
 [/INST] 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 $[0,1] \cup (16, +\infty)$.
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