Confidence Booster

$\begin{aligned} f(x) & = \frac 1{x-1} - \frac 4{x-2} + \frac 4{x-3} - \frac 1{x-4} & \small \color{#3D99F6} \text{Let }u = x-\frac 52 \\ & = \color{#3D99F6} \frac 1{u+\frac 32}{\color{#D61F06} - \frac 4{u+\frac 12} + \frac 4{u-\frac 12}} - \frac 1{u-\frac 32} \\ & = \color{#D61F06} \frac 4{u^2-\frac 14} \color{#3D99F6} - \frac 3{u^2-\frac 94} \\ & = \frac {u^2-\frac {33}4}{u^4-\frac 52 u^2+\frac 9{16}} \\ & = \frac {16u^2-132}{16u^4-40 u^2+9} \end{aligned}$

Now we have:

$\begin{aligned} \frac {16u^2-132}{16u^4-40 u^2+9} & < \frac 1{30} \\ 16u^4-40 u^2+9 & > 480u^2 - 3960 \\ 16u^4-520u^2 + 3969 & > 0 \\ (4u^2-49)(4u^2-81) & > 0 \end{aligned}$

$\implies \begin{cases} u < \frac 72 & \implies x < 6 \\ u > \frac 92 & \implies x > 7 \end{cases}$

We note that $f(x)$ is not defined when $x = 1,2,3,4$ . The acceptable $x$ are 5, 8 and 9; that is $\boxed{3}$ natural numbers.

x can't be 1; 2; 3 or 4 as the equation will go wrong. So we replace x with 5; 6; 7; 8; 9. And we shall have the final answer: 3.