solution: A real number $x$ is in the domain of $g$ if and only if $$(x-3)^2 - (x-8)^2 \ge 0.$$ Expanding this out and simplifying, we get $$10x - 55\ge 0;$$ the smallest solution is $x=\frac{55}{10}=\boxed{\frac{11}{2}}$.
  
  Alternatively, once we have the quadratic equation $$(x-3)^2 - (x-8)^2 \ge 0,$$ instead of expanding it out, we can observe that $(x-3)^2$ is the square of the distance from $x$ to $3$ on the number line, while $(x-8)^2$ is the square of the distance from $x$ to $8$. Thus, $(x-3)^2-(x-8)^2\ge 0$ is true if $x$ is closer to $8$ than to $3$, which is true if and only if $x\ge \frac{8+3}{2} = \boxed{\frac{11}{2}}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  What is the smallest real number $x$ in the domain of the function $$g(x) = \sqrt{(x-3)^2-(x-8)^2}~?$$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \frac{11}{2}
extra_info.question: What is the smallest real number $x$ in the domain of the function $$g(x) = \sqrt{(x-3)^2-(x-8)^2}~?$$
extra_info.split: test