<s>[INST] Critical points of a function $f(x) = x\sqrt{x-a}$ Find the critical points of a function $f(x) = x\sqrt{x-a}$.
A function $f(x)$ is said to have critical points at points $c$ such that $f^\prime(c)$ is $0$ or undefined.
For a function $f(x) = x\sqrt{x-a}$ :
$$\hspace{-0.5 in} f^\prime(x) = \frac{3x - 2a}{2\sqrt{x - a}}$$
It may be observed that the resulting derivative is a rational function $f^\prime(x) = \frac{g(x)}{h(x)}$ such that: 
$$
\begin{align*}
f^\prime(c) &= 0 \hspace{0.825 in} \text{ when } g(c) = 0 \text{ and } h(c) \neq 0
\\ f^\prime(c) &\text{ is undefined} \hspace{0.2 in} \text{ when } h(c) = 0
\\
\\
\\ \end{align*}
$$
By this it is found that critical points will be at values $c$ where either $g(c) = 0$ or $h(c) = 0$.
$$
\begin{align*}
g(c) = 0 \implies 3c - 2a &= 0
\\ 3c &= 2a
\\ c &= \hspace{-.0325 in}\frac{2a}{3}
\\
\\ h(c) = 0 \implies 2\sqrt{c - a} &= 0
\\ c - a &= 0
\\ c &= a
\\
\\ \end{align*}
$$
Therefore, critical points will be at values $c = \{\frac{2a}{3}, a\}$.
However, my book contradicts my answer stating $c = \frac{2a}{3} \iff a < 0$.

$\hspace{.5 in}$

My answer is for a general $a$ of any real-value and includes the $c = a$ as a critical point. What's the issue here?

Edit: 
In light of the redefined idea of a critical number being only those values $c$ which are within the domain of $f(x)$:
$$
\\ \begin{align*}
\\ x &\geq a \text{ for } a \geq 0
\\ x &\geq a \text{ for } a < 0
\\ \end{align*}
$$
Critical points must be within these domains when considering either case.
Assuming the domain of the function $f(x)$ was $(-\infty, \infty)$ there would be critical points at $x = \{ \frac{2a}{3}, a \} \text{ } \forall a$.
However, the function is only defined on the interval $[a, \infty)$ in either case.
$$
\\ \hspace{-.25 in} \frac{2a}{3} < a \text{, } \hspace{.5 in} a > 0
\\ \hspace{-.25 in} \frac{2a}{3} > a \text{, } \hspace{.5 in} a < 0
$$
This means that $x = \{ \frac{2a}{3}, a \}$ are critical points for $a < 0$, but $x = \{ a\}$, and $x \neq \{\frac{2a}{3}\}$ for $a \geq 0$.
This is not a selection, however. Insight?
 [/INST] I normally don't consider a number a critical number unless it is part of the domain of the original function. So looking at $f$ and $f'$ domain we want $x>a$. You found $x=\frac{2a}{3}$ to be a critical number. So that means we want $\frac{2a}{3}>a$  What do you get when you solve that for $a$.
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