Find the domain of defination the following functions.
`f(x)=arc cos ((2x)/(1+x))`

Answer

To find the domain of the function \( f(x) = \arccos\left(\frac{2x}{1+x}\right) \), we need to determine the values of \( x \) for which the function is defined. The function \( \arccos(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to ensure that:

\[ -1 \leq \frac{2x}{1+x} \leq 1 \]

We will break this down into two inequalities and solve them step by step.

### Step 1: Solve the first inequality \( \frac{2x}{1+x} \geq -1 \)

To solve this inequality, we first rearrange it:

\[ \frac{2x}{1+x} + 1 \geq 0 \]

This simplifies to:

\[ \frac{2x + 1 + x}{1+x} \geq 0 \]

which simplifies to:

\[ \frac{3x + 1}{1+x} \geq 0 \]

Next, we find the critical points by setting the numerator and denominator to zero:

1. \( 3x + 1 = 0 \) gives \( x = -\frac{1}{3} \)
2. \( 1 + x = 0 \) gives \( x = -1 \)

Now, we will test the intervals determined by these critical points: \( (-\infty, -1) \), \( (-1, -\frac{1}{3}) \), and \( (-\frac{1}{3}, \infty) \).

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ \frac{3(-2) + 1}{1 + (-2)} = \frac{-6 + 1}{-1} = \frac{-5}{-1} = 5 \quad (\text{positive}) \]

- For \( -1 < x < -\frac{1}{3} \) (e.g., \( x = -0.5 \)):
\[ \frac{3(-0.5) + 1}{1 + (-0.5)} = \frac{-1.5 + 1}{0.5} = \frac{-0.5}{0.5} = -1 \quad (\text{negative}) \]

- For \( x > -\frac{1}{3} \) (e.g., \( x = 0 \)):
\[ \frac{3(0) + 1}{1 + 0} = \frac{1}{1} = 1 \quad (\text{positive}) \]

Thus, the solution for the first inequality is:

\[ x \in (-\infty, -1) \cup \left[-\frac{1}{3}, \infty\right) \]

### Step 2: Solve the second inequality \( \frac{2x}{1+x} \leq 1 \)

Rearranging this inequality gives:

\[ \frac{2x}{1+x} - 1 \leq 0 \]

This simplifies to:

\[ \frac{2x - (1 + x)}{1+x} \leq 0 \]

which simplifies to:

\[ \frac{x - 1}{1+x} \leq 0 \]

Again, we find the critical points:

1. \( x - 1 = 0 \) gives \( x = 1 \)
2. \( 1 + x = 0 \) gives \( x = -1 \)

Now, we will test the intervals determined by these critical points: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ \frac{-2 - 1}{-2 + 1} = \frac{-3}{-1} = 3 \quad (\text{positive}) \]

- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1 \quad (\text{negative}) \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ \frac{2 - 1}{2 + 1} = \frac{1}{3} \quad (\text{positive}) \]

Thus, the solution for the second inequality is:

\[ x \in (-1, 1] \]

### Step 3: Find the intersection of the two solutions

Now we need to find the intersection of the two sets:

1. From the first inequality: \( (-\infty, -1) \cup \left[-\frac{1}{3}, \infty\right) \)
2. From the second inequality: \( (-1, 1] \)

The intersection is:

\[ [-\frac{1}{3}, 1] \]

### Conclusion

Thus, the domain of the function \( f(x) = \arccos\left(\frac{2x}{1+x}\right) \) is:

\[ \boxed{[-\frac{1}{3}, 1]} \]