Question

\[ f(x) = \begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\[8pt] 2, & x = 0 \end{cases} \] Which of the following is true for \( f(x) \)

• A. is continuous for all \( x \in \mathbb{R} \)
• B. has maximum value 2
• C. has neither minimum nor maximum
• D. has minimum value 2

Hint:

When checking continuity for piecewise functions, ensure to compare the limits from both sides at the boundary points.

Answer 1

The Correct Option is B

We are given the function \( f(x) \) defined piecewise. Let's examine the continuity of the function. 

Step 1: Check if the function is continuous at \( x = 0 \). For \( f(x) \) to be continuous at \( x = 0 \), the left-hand and right-hand limits must be equal to the value at \( x = 0 \). The left-hand limit: \[ \lim_{x \to 0^-} \frac{x - |x|}{x} = \lim_{x \to 0^-} \frac{x - (-x)}{x} = \lim_{x \to 0^-} \frac{2x}{x} = -2. \] The right-hand limit: \[ \lim_{x \to 0^+} \frac{x - |x|}{x} = \lim_{x \to 0^+} \frac{x - x}{x} = 0. \] Since the left-hand and right-hand limits are not equal, the function is not continuous at \( x = 0 \). 

Step 2: Analyze the maximum and minimum values. Since \( f(x) \) takes the value 2 at \( x = 0 \) and the absolute values of the function outside of 0 never exceed 2, the maximum value is 2. Thus, the function has a maximum value of 2.