Enough Information?

$\begin{aligned} f(x) & = \sqrt{x^2-2x+1} + \sqrt{x^2-4x+4} \\ & = \sqrt{(x-1)^2} + \sqrt{(x-2)^2} \\ & = |x-1| + |x-2| \end{aligned}$

$\implies \begin{cases} \text{For } x < 1 & f(x) = -x+1 -x+2 & = 3 -2x \\ \text{For } 1 \le x < 2 & f(x) = x-1 -x+2 & = 1 \\ \text{For } x \ge 2 & f(x) = x-1 + x-2 & = 2x-3 \end{cases}$

Therefore, for $\dfrac {11}{10} < x < \dfrac {19.7}{11}$ , it is within $1 \le x < 2$ , then $f(x) = \boxed{1}$ .

Recall that $\sqrt{x^2} = |x|.$

$\sqrt{x^2 - 2x + 1} = \sqrt{(x - 1)^2} = |x - 1|$

$\sqrt{x^2 - 4x + 4} = \sqrt{(x - 2)^2} = |x - 2|$

The quantity $|x - 1|$ is positive because $x$ is bounded below by $1.1.$ However, $|x - 2|$ is negative because it is never greater than $2$ , being bounded above by $1.7\overline{90}.$ So we have

$|x - 1| + |x - 2| =$

$(x - 1) + \color{#D61F06}{(2 - x)}$ $=$

$-1 + 2 =$

$1$

We can use mediant fractions here..