Periods of fractions

$\begin{aligned} f(x) & = \frac 1{2\color{#3D99F6}\{-x\}} - \{x\} & \small \color{#3D99F6} \text{Note that } \{-x\} = 1 - \{x\} \\ & = \frac 1{2\color{#3D99F6}(1-\{x\})} - \{x\} & \small \color{#3D99F6} \implies b = \lim_{\{x\} \to 1} f(x) = \infty \\ & = {\color{#3D99F6} \frac 1{2(1-\{x\})} + (1 - \{x\})} - 1 & \small \color{#3D99F6} \text{By AM-GM inequality: }\frac 1{2(1-\{x\})} + (1 - \{x\}) \ge \frac 2{\sqrt 2} \\ & \ge \sqrt 2 - 1 \end{aligned}$

$\implies a = \sqrt 2 - 1 \approx \boxed{0.414}$

Note that if $x$ is an integer then $\{-x\}$ will be equal to $0$ making $f(x)$ undefined. Thus $x$ is a non-integer and by the following property

$\{x\} + \{-x\} = \begin{cases} 0, & \text{if } x \in \mathbb{Z} \\ 1, & \text{if } x \in \mathbb{R} - \mathbb{Z} \end{cases}$

we note that $\{-x\} = 1 - \{x\}$ .

So

$\dfrac{1}{2 \{-x\}} - \{x\} = \dfrac{1}{2 \left( 1 - \{x\} \right)} - \{x\}$

For simple calculations, let us denote $\{x\}$ by $t$ and equate $f(x)$ to $y$ to get

$\dfrac{1}{2(1-t)} - t = y \implies 2t^2 + t (2y-2) + (1-2y) = 0$

As $t = \{x\}$ is real, therefore the discriminant $\Delta$ of the above quadratic in terms of $t$ is non-negative. Thus

$\begin{aligned} & \Delta \ge 0 \\ \implies & {(2y-2)}^2 - 4 \cdot 2 \cdot (1-2y) \ge 0 \\ \implies & y^2 + 2y - 1 \ge 0 \\ \implies & y \in \left( - \infty , -1 - \sqrt{2} \right] \cup \left[ -1 + \sqrt{2} , \infty \right) \end{aligned}$

But since $t \in (0,1)$ , we note that $y = \dfrac{1}{2(1-t)} - t$ is always positive in this interval. Hence

$y \in \left[ -1 + \sqrt{2} , \infty \right)$

Note that $y = f(x)$ itself is the range of $f(x)$ and thus

$a = -1 + \sqrt{2} = \boxed{0.414}$