
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


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For a function $y = f(x)$ defined on the interval $D$, if there exists $x_0 \in D$ such that for every $x \in D$, $f(x) \geq f(x_0)$, then the function $f(x)$ is said to have a "lower bound" on the interval $D$, and $f(x_0)$ is called the "lower bound" of the function $f(x)$ on $D$.
(1) Determine whether the following functions have a "lower bound". If they do, provide the "lower bound"; otherwise, explain why not: $f_1(x) = 1 - 2x$ ($x > 0$), $f_2(x) = x + \frac{16}{x}$ ($0 < x \leq 5$).
(2) By analogy with the definition of a function having a "lower bound," write out the definition for a function $f(x)$ having an "upper bound" on interval $D$, and determine whether the function $f_2(x) = |x - \frac{16}{x}|$ ($0 < x \leq 5$) has an "upper bound"? Explain your reasoning.
(3) If the function $f(x)$ has both an "upper bound" and a "lower bound" on the interval $D$, it is called a "bounded function" on $D$, and the difference between the "upper bound" and "lower bound" is called the "amplitude $M$" of the function $f(x)$ on $D$. For a real number $a$, explore whether the function $F(x) = x|x - 2a| + 3$ ($a \leq \frac{1}{2}$) is a "bounded function" on the interval $[1, 2]$. If it is, find the value of the "amplitude $M$".

Present the answer in LaTex format: \boxed{Your answer}
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