`f(x)=x+|x|` is continuous for

Options

A. `x in (-oo, oo)`
B. `x in (-oo, oo) -{0}`
C. only `x gt 0`
D. no value of x

Explanation

To determine the continuity of the function \( f(x) = x + |x| \), we will analyze it step by step.

### Step 1: Understand the Absolute Value Function
The absolute value function \( |x| \) behaves differently based on whether \( x \) is positive or negative. We need to consider two cases:
- Case 1: \( x \geq 0 \)
- Case 2: \( x < 0 \)

### Step 2: Evaluate \( f(x) \) for \( x \geq 0 \)
For \( x \geq 0 \):
\[ |x| = x \]
Thus, the function becomes:
\[ f(x) = x + |x| = x + x = 2x \]

### Step 3: Evaluate \( f(x) \) for \( x < 0 \)
For \( x < 0 \):
\[ |x| = -x \]
Thus, the function becomes:
\[ f(x) = x + |x| = x - x = 0 \]

### Step 4: Define the Piecewise Function
Now we can define the function \( f(x) \) as a piecewise function:
\[ f(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]

### Step 5: Check Continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify:
1. \( f(0) \)
2. The left-hand limit \( \lim_{x \to 0^-} f(x) \)
3. The right-hand limit \( \lim_{x \to 0^+} f(x) \)

Calculating these:
- \( f(0) = 2(0) = 0 \)
- \( \lim_{x \to 0^-} f(x) = 0 \) (since for \( x < 0 \), \( f(x) = 0 \))
- \( \lim_{x \to 0^+} f(x) = 2(0) = 0 \) (since for \( x \geq 0 \), \( f(x) = 2x \))

Since:
\[ f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 0 \]
the function is continuous at \( x = 0 \).

### Step 6: Conclusion on Continuity
Since \( f(x) \) is defined and continuous for all \( x \) in both cases (for \( x < 0 \) and \( x \geq 0 \)), we conclude that:
\[ f(x) \text{ is continuous for all } x \in (-\infty, \infty). \]

### Final Answer
The function \( f(x) = x + |x| \) is continuous for \( x \in (-\infty, \infty) \).

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