The function $ f$ has the property that for each real number $ x$ in its domain, $ 1/x$ is also in its domain and
\[ f(x) \plus{} f\left(\frac {1}{x}\right) \equal{} x.
\]What is the largest set of real numbers that can be in the domain of $ f$?
$ \textbf{(A) } \{ x | x\ne 0\} \qquad \textbf{(B) } \{ x | x < 0\} \qquad \textbf{(C) }\{ x | x > 0\}\\
\textbf{(D) } \{ x | x\ne \minus{} 1 \text{ and } x\ne 0 \text{ and } x\ne 1\} \qquad \textbf{(E) } \{ \minus{} 1,1\}$
1. Given the function \( f \) has the property that for each real number \( x \) in its domain, \( \frac{1}{x} \) is also in its domain, and it satisfies the equation:
   \[
   f(x) + f\left(\frac{1}{x}\right) = x.
   \]

2. To explore the domain of \( f \), we first substitute \( \frac{1}{x} \) into the given equation:
   \[
   f\left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.
   \]

3. Notice that the original equation and the substituted equation are symmetric:
   \[
   f(x) + f\left(\frac{1}{x}\right) = x,
   \]
   \[
   f\left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.
   \]

4. By comparing these two equations, we see that:
   \[
   x = \frac{1}{x}.
   \]

5. Solving \( x = \frac{1}{x} \):
   \[
   x^2 = 1 \implies x = \pm 1.
   \]

6. Therefore, the only values of \( x \) that satisfy both equations are \( x = 1 \) and \( x = -1 \).

7. Hence, the largest set of real numbers that can be in the domain of \( f \) is \( \{-1, 1\} \).

The final answer is \(\boxed{\{-1, 1\}}\).