For what values a and b `f(x)={{:(,(x+2)/(|x+2|)+a,if x lt -2),(,a+b,if x=-2),(,(x+2)/(|x+2|)+2b, if x gt -2):}` Is continuous at x=-2

Answer

To determine the values of \( a \) and \( b \) for which the function

\[ f(x) = \begin{cases} \frac{x+2}{|x+2|} + a & \text{if } x < -2 \\ a + b & \text{if } x = -2 \\ \frac{x+2}{|x+2|} + 2b & \text{if } x > -2 \end{cases} \]

is continuous at \( x = -2 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at \( x = -2 \) are all equal.

### Step 1: Calculate the left-hand limit as \( x \) approaches -2

The left-hand limit is given by:

\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \left( \frac{x+2}{|x+2|} + a \right) \]

For \( x < -2 \), \( |x+2| = -(x+2) \), so we have:

\[ \frac{x+2}{|x+2|} = \frac{x+2}{-(x+2)} = -1 \]

Thus, the left-hand limit becomes:

\[ \lim_{x \to -2^-} f(x) = -1 + a \]

### Step 2: Calculate the right-hand limit as \( x \) approaches -2

The right-hand limit is given by:

\[ \lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} \left( \frac{x+2}{|x+2|} + 2b \right) \]

For \( x > -2 \), \( |x+2| = x+2 \), so we have:

\[ \frac{x+2}{|x+2|} = \frac{x+2}{x+2} = 1 \]

Thus, the right-hand limit becomes:

\[ \lim_{x \to -2^+} f(x) = 1 + 2b \]

### Step 3: Evaluate the function at \( x = -2 \)

At \( x = -2 \), the function value is:

\[ f(-2) = a + b \]

### Step 4: Set the limits and function value equal

For \( f(x) \) to be continuous at \( x = -2 \), we need:

\[ \text{Left-hand limit} = \text{Right-hand limit} = f(-2) \]

This gives us the following equations:

1. \( -1 + a = 1 + 2b \)
2. \( -1 + a = a + b \)

### Step 5: Solve the equations

From equation (2):

\[ -1 + a = a + b \implies -1 = b \implies b = -1 \]

Substituting \( b = -1 \) into equation (1):

\[ -1 + a = 1 + 2(-1) \implies -1 + a = 1 - 2 \implies -1 + a = -1 \implies a = 0 \]

### Conclusion

The values of \( a \) and \( b \) for which the function is continuous at \( x = -2 \) are:

\[ \boxed{a = 0, b = -1} \]