Let's solve the problem step by step.

Problem Statement Recap:

1. The function $f(x)$ is defined by: $f(x) = \frac{1 - x}{1 + x}, \quad x \neq -1.$

2. First part: Show that $f(f(x)) = x$.

3. Second part: Write down $f^{-1}(x)$, the inverse of $f(x)$.

Solution:

Part 1: Showing that $f(f(x)) = x$

Given $f(x) = \frac{1-x}{1+x}$, let's compute $f(f(x))$:

1. Substitute $f(x)$ into itself: $f(f(x)) = f\left(\frac{1-x}{1+x}\right)$

2. Now, replace $x$ in the original function with $\frac{1-x}{1+x}$: $f\left(\frac{1-x}{1+x}\right) = \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}}$

3. Simplify the expression step by step: $1 - \frac{1-x}{1+x} = \frac{(1+x) - (1-x)}{1+x} = \frac{1+x - 1 + x}{1+x} = \frac{2x}{1+x}$ $1 + \frac{1-x}{1+x} = \frac{(1+x) + (1-x)}{1+x} = \frac{1+x + 1-x}{1+x} = \frac{2}{1+x}$

   So, the expression for $f(f(x))$ becomes: $f(f(x)) = \frac{\frac{2x}{1+x}}{\frac{2}{1+x}} = x$

Thus, we have shown that $f(f(x)) = x$.

Part 2: Finding the Inverse $f^{-1}(x)$

To find the inverse function $f^{-1}(x)$, we need to solve the equation $y = f(x)$ for $x$:

1. Start with: $y = \frac{1 - x}{1 + x}$

2. Cross-multiply to clear the fraction: $y(1 + x) = 1 - x$ $y + yx = 1 - x$

3. Collect terms involving $x$ on one side: $yx + x = 1 - y$ $x(y + 1) = 1 - y$

4. Solve for $x$: $x = \frac{1 - y}{y + 1}$

So, the inverse function $f^{-1}(x)$ is: $f^{-1}(x) = \frac{1 - x}{1 + x}$

Summary:

• First part: $f(f(x)) = x$.
• Second part: The inverse function $f^{-1}(x)$ is $f^{-1}(x) = \frac{1 - x}{1 + x}$.

Would you like further details or have any questions?

Related Questions:

1. How do you determine if a function is bijective?
2. What are the necessary steps to prove that a function is invertible?
3. Can $f^{-1}(x)$ have the same form as $f(x)$ for all functions?
4. How does the condition $x \neq -1$ affect the domain and range of $f(x)$?
5. What are the implications of a function having $f(f(x)) = x$?

Tip:

When dealing with functions and their inverses, always check if the composition of the function with its supposed inverse yields the identity function $x$.