14.3 If $\frac{6}{b}<x<\frac{10}{b}$, find the value of $c=\sqrt{x^{2}-2 x+1}+\sqrt{x^{2}-6 x+9}$.
If $\frac{6}{b}<x<\frac{10}{b}$, determine the value of $c=\sqrt{x^{2}-2 x+1}+\sqrt{x^{2}-6 x+9}$.
$$
1.5<x<2.5, c=\sqrt{(x-1)^{2}}+\sqrt{(x-3)^{2}}=x-1+3-x=2
$$

Remark: original question: ……the greatest possible value of $c$.
The value of $c$ is a constant.