Question

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

• A. \( x \in (-\infty, \infty) \)
• B. \( x \in (-\infty, \infty) - \{0\} \)
• C. only \( x > 0 \)
• D. No value for x

Hint:

Functions involving absolute values are often best analyzed by converting them into piecewise functions. The sum of two continuous functions is always continuous. Since \( g(x)=x \) is continuous and \( h(x)=|x| \) is continuous, their sum \( f(x)=g(x)+h(x) \) must also be continuous everywhere.

Answer 1

The Correct Option is A

We can write the function \( f(x) = x + |x| \) as a piecewise function by considering the definition of \( |x| \). \[ |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} \] 

So, \[ f(x) = \begin{cases} x + x = 2x, & x \geq 0 \\ x + (-x) = 0, & x < 0 \end{cases} \] 

The function is composed of two linear pieces, \( y=2x \) and \( y=0 \). Linear functions are continuous everywhere. The only point where continuity might be an issue is at the "join" point, \( x=0 \). To check for continuity at \( x=0 \), we check the left-hand limit, the right-hand limit, and the function value. \[\begin{array}{rl} \bullet & \text{Function value: \( f(0) = 2(0) = 0 \).} \\ \bullet & \text{Left-hand limit (LHL): \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 0 = 0 \).} \\ \bullet & \text{Right-hand limit (RHL): \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x = 2(0) = 0 \).} \\ \end{array}\] Since LHL = RHL = \( f(0) \), the function is continuous at \( x=0 \). Because it is continuous for \( x < 0 \), continuous for \( x > 0 \), and continuous at \( x=0 \), the function is continuous for all real numbers, i.e., \( x \in (-\infty, \infty) \).