6. 54 Find the value of 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, so $f(0) = \operatorname{arctg} 2 + c = 0$. From this, 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
$$f(x) = \operatorname{arctg} \frac{2-2x}{1+4x} - \operatorname{arctg} 2$$
is an odd function, i.e., it satisfies the relation:
$$f(x) = -f(-x).$$

In fact, when $x \in \left(-\frac{1}{4}, \frac{1}{4}\right)$, the value of the function $z = \frac{2-2x}{1+4x}$ falls 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 formulas, we get
$$\begin{array}{l}
\operatorname{tg} f(x) = -2x \\
\operatorname{tg}[-f(-x)] = -2x
\end{array}$$

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