Valley and Mountain

$\begin{aligned} f(x) & = \tan \left(\frac {\pi x}{x^2+4} \right) \\ \implies \frac d{dx} f(x) & = \frac {\pi \left(x^2+4\right) - 2 \pi x^2}{\left(x^2+4\right)^2} \cdot \sec^2 \left(\frac {\pi x}{x^2+4} \right) \\ f'(x) & = \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \cdot \sec^2 \left(\frac {\pi x}{x^2+4} \right) \end{aligned}$

Making $f'(x) = 0$ to find the maximum and minimum, we note that since $\sec^2 \left(\dfrac {\pi x}{x^2+4} \right) \ne 0$ $\implies 4 - x^2 = 0$ , $\implies x = \pm 2$ . Let us check the values of $f''(2)$ and $f''(-2)$ to determine which is maximum and minimum.

$\begin{aligned} f'(x) & = \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \cdot \sec^2 \left(\frac {\pi x}{x^2+4} \right) \\ & = \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \left(1 + \tan^2 \left(\frac {\pi x}{x^2+4} \right) \right) \\ & = \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \left(1 + f^2(x) \right) \\ f''(x) & = \frac {\pi \left(- 2x\left(x^2+4\right)-\left(4 - x^2\right)(2x)\right)}{\left(x^2+4\right)^3} \left(1 + f^2(x) \right) + \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \left(1 + 2f(x)f'(x) \right) \\ & = \frac {-8\pi x}{\left(x^2+4\right)^3} \left(1 + f^2(x) \right) + \frac {\pi \left(4 - x^2\right)}{\left(x^2+4\right)^2} \left(1 + 2f(x)f'(x) \right) \end{aligned}$

$\implies \begin{cases} f''(2) < 0 & \implies f(2) = \tan \left(\dfrac {2 \pi }{2^2+4} \right) = 1 & \text{Maximum} \\ f''(-2) > 0 & \implies f(-2) = \tan \left(\dfrac {-2 \pi }{2^2+4} \right) = -1 & \text{Minimum} \end{cases}$

The length of the line joining $(2,1)$ and $(-2,-1)$ , $L = \sqrt{(2+2)^2+(1+1)^2} = \sqrt{20} = \boxed{2\sqrt 5}$ .