Okay, let's try to figure out the smallest real number x in the domain of the function g(x) = sqrt[(x-3)^2 - (x-8)^2]. Hmm, so the domain of a square root function requires that the expression inside the square root is non-negative. That means (x-3)^2 - (x-8)^2 must be greater than or equal to zero. Right?

First, maybe I should simplify the expression inside the square root. Let's expand both squares. Let me compute (x-3)^2 and (x-8)^2 separately. 

(x - 3)^2 = x^2 - 6x + 9
(x - 8)^2 = x^2 - 16x + 64

Now, subtract the second expansion from the first: (x^2 - 6x + 9) - (x^2 - 16x + 64). Let's do the subtraction term by term.

x^2 - x^2 cancels out. Then -6x - (-16x) is -6x +16x = 10x. Then 9 - 64 is -55. So the expression inside the square root simplifies to 10x -55. 

Therefore, the inequality we need to solve is 10x -55 ≥ 0. Let's solve for x. Adding 55 to both sides gives 10x ≥ 55. Dividing both sides by 10 gives x ≥ 5.5. 

Wait, so does that mean the domain is [5.5, ∞)? Then the smallest real number x in the domain would be 5.5. But let me check if I did the algebra correctly. Let me verify the expansion again.

First term: (x - 3)^2 = x² -6x +9. Correct. Second term: (x -8)^2 = x² -16x +64. Correct. Subtracting the second from the first: x² -6x +9 -x² +16x -64. The x² terms cancel. Then -6x +16x is 10x. 9 -64 is -55. So yes, 10x -55. Then set ≥0. 10x ≥55. x≥5.5. 

Hmm, 5.5 is 11/2. So 5.5 is the smallest x. But wait, let me confirm with another approach. Maybe factor the original expression.

Original expression inside sqrt: (x-3)^2 - (x-8)^2. That's a difference of squares, which factors as [ (x-3) - (x-8) ] [ (x-3) + (x-8) ]. Let's compute each bracket.

First bracket: (x -3) - (x -8) = x -3 -x +8 = 5.

Second bracket: (x -3) + (x -8) = 2x -11.

So the expression factors to 5*(2x -11). Therefore, the expression inside the sqrt is 5*(2x -11). So 5*(2x -11) must be ≥0.

Since 5 is positive, we can divide both sides by 5 without changing the inequality. So 2x -11 ≥0. Then 2x ≥11, so x ≥11/2, which is 5.5. Same result. So that's correct.

Therefore, the domain is [5.5, ∞), so the smallest real number x is 5.5. 

But let me check with another value. Let's pick x=5.5 and plug into the original expression.

(x-3)^2 - (x-8)^2 when x=5.5:

(5.5 -3)^2 - (5.5 -8)^2 = (2.5)^2 - (-2.5)^2 = 6.25 -6.25 =0. So sqrt(0)=0, which is valid.

If x is less than 5.5, say x=5, then:

(5 -3)^2 - (5 -8)^2 = (2)^2 - (-3)^2 =4 -9= -5. Which is negative, so sqrt(-5) is not real. So indeed, x must be at least 5.5.

Therefore, the smallest real number x in the domain is 5.5, which is 11/2. 

Wait, just to make sure, what if x is a real number just above 5.5, like 5.6? Let's compute:

(5.6 -3)^2 - (5.6 -8)^2 = (2.6)^2 - (-2.4)^2 =6.76 -5.76=1. So sqrt(1)=1, which is valid. So x=5.6 is allowed.

And x=5.5 is allowed as we saw. So yes, 5.5 is the boundary. 

Alternatively, maybe someone might think that the domain is over complex numbers, but the question says "smallest real number in the domain", so we only consider real numbers. So complex numbers are out of the question here.

Therefore, after all these checks, I'm confident that the smallest real number x in the domain is 5.5, which is 11/2. So the answer is 11/2, or 5.5. 

But since the problem asks for the answer in boxed notation, probably as a fraction. 11/2 is 5.5, so either is acceptable, but fractions are often preferred in mathematical contexts. So I'll go with 11/2.

**Final Answer**
The smallest real number \( x \) in the domain of the function is \boxed{\frac{11}{2}}.
Therefore, the smallest real number $x$ in the domain of the function is $\boxed{\frac{11}{2}}$.
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