Let `f:[-sqrt(2)+1,sqrt(2)+1]to[(-sqrt(2)+1),2,(sqrt(2)+1)/2]` be a function defined by `f(x)=(1-x)/(1+x^(2))`.
If `f^(-1)(x)={((-1+lamda(sqrt(4x-4x^(2)+1)))/(2x),x!=0),(mu,x=0):}` then `lamda+mu` is.

Explanation

To solve the problem step by step, we need to find the inverse of the function \( f(x) = \frac{1-x}{1+x^2} \) and determine the values of \( \lambda \) and \( \mu \) in the expression for \( f^{-1}(x) \).

### Step 1: Set up the equation
We start with the function:
\[ y = \frac{1 - x}{1 + x^2} \]
We need to express \( x \) in terms of \( y \). Rearranging gives:
\[ y(1 + x^2) = 1 - x \]
This simplifies to:
\[ yx^2 + xy + y - 1 = 0 \]

### Step 2: Use the quadratic formula
This is a quadratic equation in \( x \):
\[ yx^2 + xy + (y - 1) = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify \( a = y \), \( b = y \), and \( c = y - 1 \):
\[ x = \frac{-y \pm \sqrt{y^2 - 4y(y - 1)}}{2y} \]

### Step 3: Simplify the expression under the square root
Calculating the discriminant:
\[ y^2 - 4y^2 + 4y = -3y^2 + 4y \]
Thus, we have:
\[ x = \frac{-y \pm \sqrt{-3y^2 + 4y}}{2y} \]

### Step 4: Choose the correct sign
According to the problem, we will take the positive root:
\[ x = \frac{-y + \sqrt{-3y^2 + 4y}}{2y} \]
This gives us the inverse function:
\[ f^{-1}(y) = \frac{-y + \sqrt{4y - 4y^2 + 1}}{2y} \]

### Step 5: Identify \( \lambda \) and \( \mu \)
From the problem, we have:
\[ f^{-1}(x) = \left\{ \frac{-1 + \lambda \sqrt{4x - 4x^2 + 1}}{2x}, x \neq 0 \right\}, \quad f^{-1}(0) = \mu \]
We can compare:
\[ \frac{-y + \sqrt{4y - 4y^2 + 1}}{2y} = \frac{-1 + \lambda \sqrt{4x - 4x^2 + 1}}{2x} \]
From this, we can identify that \( \lambda = 1 \).

### Step 6: Calculate \( f^{-1}(0) \)
To find \( \mu \), we need to evaluate \( f^{-1}(0) \):
\[ f^{-1}(0) = \lim_{x \to 0} \frac{-x + \sqrt{4x - 4x^2 + 1}}{2x} \]
Using L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{-1 + \frac{4 - 8x}{2\sqrt{4x - 4x^2 + 1}}}{2} \]
Evaluating this limit as \( x \to 0 \):
\[ = \frac{-1 + 2}{2} = \frac{1}{2} \]
Thus, \( \mu = 1 \).

### Step 7: Calculate \( \lambda + \mu \)
Now we can find:
\[ \lambda + \mu = 1 + 1 = 2 \]

### Final Answer
\[ \lambda + \mu = 2 \]