<Question>
What is the smallest real number $x$ in the domain of the function $$g(x) = \sqrt{(x-3)^2-(x-8)^2}~?$$
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest real number x in the domain of the function g(x) = sqrt[(x-3)^2 - (x-8)^2]. Hmm, let's start by recalling that the domain of a square root function is all the real numbers for which the expression inside the square root is non-negative. That means I need to solve the inequality (x-3)^2 - (x-8)^2 ≥ 0.

First, let me write down the inequality:

(x - 3)^2 - (x - 8)^2 ≥ 0

I remember that there's a difference of squares formula which says a^2 - b^2 = (a - b)(a + b).
Maybe I can apply that here. Let me let a = x - 3 and b = x - 8. Then the expression becomes:

(a - b)(a + b) ≥ 0

Let me compute a - b and a + b.

<prompt_1.1> Apply the difference of squares formula to the expression and compute \(a - b\) and \(a + b\). </prompt_1.1>
-----

Alternatively, I could expand both squares and subtract. Let me try that method to confirm:

<prompt_1.2> Expand both squares and subtract to confirm the inequality. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
a - b = (x - 3) - (x - 8) = x - 3 - x + 8 = (x - x) + (-3 + 8) = 0 + 5 = 5

a + b = (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11

So the expression becomes:

5 * (2x - 11) ≥ 0
</execution_1.1>

<execution_1.2>
First, expand (x - 3)^2:

(x - 3)^2 = x^2 - 6x + 9

Then expand (x - 8)^2:

(x - 8)^2 = x^2 - 16x + 64

Subtract the second expansion from the first:

(x^2 - 6x + 9) - (x^2 - 16x + 64) = x^2 -6x +9 -x^2 +16x -64

Simplify:

The x^2 terms cancel out. Then -6x +16x = 10x. Then 9 -64 = -55.

So we get 10x -55 ≥ 0, which is the same as before.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, solving for x:

5*(2x - 11) ≥ 0

Since 5 is a positive constant, multiplying both sides by 5 won't change the inequality direction. So divide both sides by 5:

<prompt_2.1> Solve the inequality \(5*(2x - 11) \geq 0\). </prompt_2.1>
-----

Based on execution_1.2:
So, solving 10x -55 ≥ 0:

<prompt_2.2> Reconfirm the solution by solving \(10x - 55 \geq 0\). </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
2x - 11 ≥ 0

Then, add 11 to both sides:

2x ≥ 11

Divide both sides by 2:

x ≥ 11/2

So, 11/2 is 5.5. Therefore, the domain of g(x) is all real numbers x such that x is greater than or equal to 5.5. Therefore, the smallest real number x in the domain is 5.5, or 11/2.
</execution_2.1>

<execution_2.2>
10x ≥ 55

x ≥ 55/10

Simplify 55/10: divide numerator and denominator by 5: 11/2. So 11/2 is 5.5. So x ≥ 5.5.

Therefore, the domain of g(x) is [5.5, ∞), so the smallest real number x in the domain is 5.5.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, let me check if I did everything correctly. Let me verify the steps again.

<prompt_3.1> Verify the solution by checking the steps again. </prompt_3.1>
-----

But wait, hold on. Let me make sure there are no mistakes in the algebra. Let me re-express the original inequality:

<prompt_3.2> Re-express the original inequality and confirm the solution. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Starting with (x - 3)^2 - (x - 8)^2 ≥ 0. Applied difference of squares: (a^2 - b^2) = (a - b)(a + b). Then a - b was 5, and a + b was 2x - 11. Multiplying them gives 5*(2x -11) ≥ 0. Then solving for x, we get x ≥ 11/2. That seems correct.
</execution_3.1>

<execution_3.2>
(x - 3)^2 - (x - 8)^2 ≥ 0

Expanding both terms:

(x^2 - 6x + 9) - (x^2 - 16x + 64) ≥ 0

Simplify:

x^2 -6x +9 -x^2 +16x -64 ≥ 0

The x^2 terms cancel:

(-6x +16x) + (9 -64) ≥ 0

10x -55 ≥ 0

Yes, that's correct. So 10x ≥ 55 => x ≥ 5.5.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_3.2:
But let me think again. Since this is a square root function, the expression inside must be non-negative.
So, we have to ensure that (x - 3)^2 - (x - 8)^2 is greater than or equal to zero.

But perhaps I can also analyze the expression (x - 3)^2 - (x - 8)^2 as a quadratic function and see where it's non-negative.
Let me denote f(x) = (x - 3)^2 - (x - 8)^2. From earlier, we found that f(x) = 10x -55. So f(x) is a linear function with slope 10, which is positive, so it's increasing. The root of the function is at x = 55/10 = 5.5. So for x ≥ 5.5, f(x) ≥ 0, and for x < 5.5, f(x) < 0. Therefore, the domain is [5.5, ∞).
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the smallest real number $x$ in the domain of $g$ is $\boxed{\frac{11}{2}}$.
</Final_answer>
</think>