Understanding the Function \(f(x) = x + |x|\)

We are asked to determine the interval over which the function defined as \(f(x) = x + |x|\) is continuous. To do this, we first need to understand the behavior of the absolute value function, $|x|$.

The absolute value function $|x|$ is defined piecewise:

• \(|x| = x\) if \(x \ge 0\)
• \(|x| = -x\) if \(x < 0\)

Using this definition, we can rewrite the function $f(x)$ as a piecewise function:

This simplification comes from substituting the definition of $|x|$ into the original function expression.

Analyzing Continuity in Intervals

A function is continuous over an interval if it is continuous at every point within that interval. We examine the continuity of $f(x)$ in the intervals $x < 0$, $x > 0$, and at the point $x = 0$ where the function definition changes.

Continuity for $x < 0$

For $x < 0$, the function is defined as \(f(x) = 0\). This is a constant function. Constant functions are always continuous everywhere. Therefore, $f(x)$ is continuous for all $x < 0$. The limit as $x$ approaches any value \(c < 0\) is \(\lim_{x \to c} 0 = 0\), and $f(c) = 0$. Thus, the function is continuous in the interval \( (-\∞, 0) \).

Continuity for $x > 0$

For $x > 0$, the function is defined as \(f(x) = 2x\). This is a linear function (a polynomial of degree 1). Polynomial functions are continuous everywhere. Therefore, $f(x)$ is continuous for all $x > 0$. The limit as $x$ approaches any value \(c > 0\) is \(\lim_{x \to c} 2x = 2c\), and $f(c) = 2c$. Thus, the function is continuous in the interval \( (0, \∞) \).

Checking Continuity at $x = 0$

To determine if the function is continuous over all real numbers, we must check continuity at the point $x = 0$, where the definition of the function changes. For a function to be continuous at a point $c$, three conditions must be met:

1. \(f(c)\) must be defined.
2. The limit \(\lim_{x \to c} f(x)\) must exist.
3. The limit must equal the function value: \(\lim_{x \to c} f(x) = f(c)\).

Let's check these conditions for $x = 0$:

1. Is \(f(0)\) defined? According to the piecewise definition, for $x \ge 0$, $f(x) = 2x$. So, $f(0) = 2 \times 0 = 0$. Yes, $f(0)$ is defined.
2. Does \(\lim_{x \to 0} f(x)\) exist? For the limit to exist, the left-hand limit and the right-hand limit must be equal.
   • Left-hand limit ($x \to 0^-$): As $x$ approaches 0 from the negative side ($x < 0$), $f(x) = 0$. \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 0 = 0 \]
   • Right-hand limit ($x \to 0^+$): As $x$ approaches 0 from the positive side ($x > 0$), $f(x) = 2x$. \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x = 2 \times 0 = 0 \]
   Since the left-hand limit ($0$) is equal to the right-hand limit ($0$), the limit exists and \(\lim_{x \to 0} f(x) = 0\).
3. Does \(\lim_{x \to 0} f(x) = f(0)\)? We found $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$. Since both are equal, the third condition is met.

Because all three conditions are satisfied, the function $f(x) = x + |x|$ is continuous at $x = 0$.

Conclusion on Continuity

Since $f(x)$ is continuous for $x < 0$, for $x > 0$, and also at $x = 0$, it is continuous for all real numbers. This means the function is continuous over the entire interval $x \in (-\∞, ∞)$.

Comparing this result with the given options:

• Option 1: $x \in (-\∞, ∞)$ - This matches our finding.
• Option 2: $x \in (-\∞, ∞) - \{0\}$ - This incorrectly excludes $x=0$.
• Option 3: only $x > 0$ - This is incomplete.
• Option 4: No value of $x$ - This is incorrect.

Therefore, the correct option is the first one, stating that the function is continuous for $x \in (-\∞, ∞)$.