Question

Let \( f(x) = |1 - 2x| \), then:

• A. \( f(x) \) is continuous but not differentiable at \( x = \frac{1}{2} \).
• B. \( f(x) \) is differentiable but not continuous at \( x = \frac{1}{2} \).
• C. \( f(x) \) is both continuous and differentiable at \( x = \frac{1}{2} \).
• D. \( f(x) \) is neither differentiable nor continuous at \( x = \frac{1}{2} \).

Hint:

Absolute value functions often introduce "corners" or sharp turns at the point where the expression inside the absolute value becomes zero. These points are typically where the function is continuous but not differentiable due to the different slopes from the left and right.

Answer 1

The Correct Option is A

Step 1: Understand the function \( f(x) = |1 - 2x| \).
The absolute value function \( |u| \) is defined as: \[ |u| = \begin{cases} u, & \text{if } u \ge 0
-u, & \text{if } u<0 \end{cases} \] Applying this to \( f(x) = |1 - 2x| \): \[ f(x) = \begin{cases} 1 - 2x, & \text{if } 1 - 2x \ge 0 \implies x \le \frac{1}{2}
-(1 - 2x) = 2x - 1, & \text{if } 1 - 2x<0 \implies x>\frac{1}{2} \end{cases} \]
Step 2: Check for continuity at \( x = \frac{1}{2} \).
For a function to be continuous at \( x = a \), we need \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).
At \( x = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = \left|1 - 2\left(\frac{1}{2}\right)\right| = |1 - 1| = 0 \). Left-hand limit: \[ \lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^-} (1 - 2x) = 1 - 2\left(\frac{1}{2}\right) = 1 - 1 = 0 \] Right-hand limit: \[ \lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}^+} (2x - 1) = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0 \] Since \( \lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^+} f(x) = f\left(\frac{1}{2}\right) = 0 \), the function \( f(x) \) is continuous at \( x = \frac{1}{2} \).
Step 3: Check for differentiability at \( x = \frac{1}{2} \).
For a function to be differentiable at \( x = a \), the left-hand derivative must be equal to the right-hand derivative at that point.
We need to find \( f'(x) \) for \( x<\frac{1}{2} \) and \( x>\frac{1}{2} \).
For \( x<\frac{1}{2} \), \( f(x) = 1 - 2x \), so \( f'(x) = -2 \).
For \( x>\frac{1}{2} \), \( f(x) = 2x - 1 \), so \( f'(x) = 2 \).
Left-hand derivative at \( x = \frac{1}{2} \): \[ f'\left(\frac{1}{2}^-\right) = \lim_{h \to 0^-} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h} = \lim_{h \to 0^-} \frac{(1 - 2(\frac{1}{2} + h)) - 0}{h} = \lim_{h \to 0^-} \frac{1 - 1 - 2h}{h} = \lim_{h \to 0^-} \frac{-2h}{h} = -2 \] Right-hand derivative at \( x = \frac{1}{2} \): \[ f'\left(\frac{1}{2}^+\right) = \lim_{h \to 0^+} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h} = \lim_{h \to 0^+} \frac{(2(\frac{1}{2} + h) - 1) - 0}{h} = \lim_{h \to 0^+} \frac{1 + 2h - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2 \] Since \( f'\left(\frac{1}{2}^-\right) = -2 \) and \( f'\left(\frac{1}{2}^+\right) = 2 \), the left-hand derivative is not equal to the right-hand derivative at \( x = \frac{1}{2} \). Therefore, \( f(x) \) is not differentiable at \( x = \frac{1}{2} \).