Find sum of all possible values of the real parameter 'b' if the difference between the largest and smallest values of the function `f (x) = x ^(2) -2bx + 1` in the interval `[0,1]` is 4.

Answer

To solve the problem, we need to find the sum of all possible values of the real parameter 'b' such that the difference between the largest and smallest values of the function \( f(x) = x^2 - 2bx + 1 \) in the interval \([0, 1]\) is 4.

### Step 1: Rewrite the function
We can rewrite the function \( f(x) \) in a different form:
\[ f(x) = (x - b)^2 + (1 - b^2) \]
This shows that the function is a parabola that opens upwards, with its vertex at \( x = b \).

### Step 2: Identify the vertex
The vertex of the parabola \( f(x) \) is at \( x = b \). We need to analyze the behavior of the function in the interval \([0, 1]\) depending on the value of \( b \).

### Step 3: Analyze cases based on the value of \( b \)
We will consider three cases based on the value of \( b \):
1. **Case 1**: \( b < 0 \)
2. **Case 2**: \( 0 \leq b \leq 1 \)
3. **Case 3**: \( b > 1 \)

### Step 4: Calculate the maximum and minimum values of \( f(x) \)

#### Case 1: \( b < 0 \)
In this case, the vertex \( b \) is outside the interval \([0, 1]\). The maximum value occurs at \( x = 1 \) and the minimum at \( x = 0 \):
\[ f(0) = 1 \quad \text{and} \quad f(1) = 1 - 2b + 1 = 2 - 2b \]
The difference is:
\[ (2 - 2b) - 1 = 1 - 2b \]
Setting this equal to 4:
\[ 1 - 2b = 4 \implies -2b = 3 \implies b = -\frac{3}{2} \]

#### Case 2: \( 0 \leq b \leq 1 \)
Here, the vertex \( b \) is within the interval. The maximum value occurs at \( x = 1 \) and the minimum at \( x = b \):
\[ f(1) = 2 - 2b \quad \text{and} \quad f(b) = (b - b)^2 + (1 - b^2) = 1 - b^2 \]
The difference is:
\[ (2 - 2b) - (1 - b^2) = 1 - 2b + b^2 \]
Setting this equal to 4:
\[ 1 - 2b + b^2 = 4 \implies b^2 - 2b - 3 = 0 \]
Factoring gives:
\[ (b - 3)(b + 1) = 0 \implies b = 3 \quad \text{or} \quad b = -1 \]
Since \( b \) must be in the interval \([0, 1]\), there are no valid solutions from this case.

#### Case 3: \( b > 1 \)
In this case, the vertex \( b \) is again outside the interval. The maximum value occurs at \( x = 1 \) and the minimum at \( x = 0 \):
\[ f(0) = 1 \quad \text{and} \quad f(1) = 2 - 2b \]
The difference is:
\[ (2 - 2b) - 1 = 1 - 2b \]
Setting this equal to 4:
\[ 1 - 2b = 4 \implies -2b = 3 \implies b = -\frac{3}{2} \]
This value is not valid in this case.

### Step 5: Collect valid values of \( b \)
The valid values of \( b \) are:
- From Case 1: \( b = -\frac{3}{2} \)
- From Case 3: \( b = \frac{5}{2} \)

### Step 6: Calculate the sum of all possible values of \( b \)
Now, we sum the valid values:
\[ -\frac{3}{2} + \frac{5}{2} = \frac{2}{2} = 1 \]

### Final Answer
The sum of all possible values of the real parameter \( b \) is:
\[ \boxed{1} \]