Find the values of a and b sucht that the function f defined by
`f(x) = {{:((x-4)/(|x-4|)+a, if x lt 4),(a+b,if x =4),((x-4)/(|x-4|)+b, if x gt 4):}`
is a continous function at `x = 4` .

Explanation

To find the values of \( a \) and \( b \) such that the function \( f \) is continuous at \( x = 4 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at \( x = 4 \) are equal.

The function is defined as:
\[ f(x) = \begin{cases} \frac{x-4}{|x-4|} + a & \text{if } x < 4 \\ a + b & \text{if } x = 4 \\ \frac{x-4}{|x-4|} + b & \text{if } x > 4 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit as \( x \) approaches 4

For \( x < 4 \):
\[ f(x) = \frac{x-4}{|x-4|} + a \]
Since \( x < 4 \), \( |x-4| = -(x-4) \), thus:
\[ f(x) = \frac{x-4}{-(x-4)} + a = -1 + a \]
Now, we find the left-hand limit:
\[ \lim_{x \to 4^-} f(x) = -1 + a \]

### Step 2: Calculate the Right-Hand Limit as \( x \) approaches 4

For \( x > 4 \):
\[ f(x) = \frac{x-4}{|x-4|} + b \]
Since \( x > 4 \), \( |x-4| = x-4 \), thus:
\[ f(x) = \frac{x-4}{x-4} + b = 1 + b \]
Now, we find the right-hand limit:
\[ \lim_{x \to 4^+} f(x) = 1 + b \]

### Step 3: Value of the Function at \( x = 4 \)

At \( x = 4 \):
\[ f(4) = a + b \]

### Step 4: Set Up the Continuity Conditions

For \( f \) to be continuous at \( x = 4 \), we need:
\[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \]
This gives us the equations:
1. \( -1 + a = 1 + b \)
2. \( 1 + b = a + b \)

### Step 5: Solve the Equations

From the first equation:
\[ -1 + a = 1 + b \implies a - b = 2 \quad \text{(Equation 1)} \]

From the second equation:
\[ 1 + b = a + b \implies 1 = a \quad \text{(Equation 2)} \]

### Step 6: Substitute and Find \( b \)

Substituting \( a = 1 \) into Equation 1:
\[ 1 - b = 2 \implies -b = 1 \implies b = -1 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:
\[ a = 1, \quad b = -1 \]

### Summary

The function \( f(x) \) is continuous at \( x = 4 \) when:
\[ \boxed{a = 1 \text{ and } b = -1} \]