$6 \cdot 54$ Find the constant $c$ such that the function
$$
f(x)=\operatorname{arctg} \frac{2-2 x}{1+4 x}+c
$$

is an odd function on the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$.
[Solution] Suppose the constant $c$ exists, and the function $f(x)$ is an odd function. Therefore, $f(0) = \operatorname{arctg} 2 + c = 0$, which implies that the only possible value for $c$ is $-\operatorname{arctg} 2$.
We now prove that in the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$, the function
$$
f(x) = \operatorname{arctg} \frac{2-2 x}{1+4 x} - \operatorname{arctg} 2
$$

is an odd function, i.e., it satisfies the relation:
$$
f(x) = -f(-x).
$$

Indeed, when $x \in \left(-\frac{1}{4}, \frac{1}{4}\right)$, the value of the function $z = \frac{2-2 x}{1+4 x}$ lies within the interval $\left(\frac{3}{4}, \infty\right)$, and thus the value of the function $f(x) = \operatorname{arctg} z - \operatorname{arctg} 2$ lies within the interval $\left(\operatorname{arctg} \frac{3}{4} - \operatorname{arctg} 2, \frac{\pi}{2} - \operatorname{arctg} 2\right)$, which is clearly within the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Therefore, when $x \in \left(-\frac{1}{4}, \frac{1}{4}\right)$, the value of the function $-f(-x)$ also lies within this interval, so equation (1) is equivalent to
$$
\operatorname{tg} f(x) = \operatorname{tg}(-f(-x)).
$$

However, using trigonometric identities, we can obtain
$$
\begin{array}{l}
\operatorname{tg} f(x) = -2 x, \\
\operatorname{tg}[-f(-x)] = -2 x,
\end{array}
$$

Thus, equation (2) holds, and consequently, equation (1) holds. Therefore, $c = -\operatorname{arctg} 2$.