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: Define the function \( f(x) = |1 - 2x| \).
The absolute value function \( |u| \) is defined as follows: \[ |u| = \begin{cases} u, & \text{if } u \ge 0
-u, & \text{if } u<0 \end{cases} \] 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 continuity at \( x = \frac{1}{2} \).
A function is continuous at \( x = a \) if \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).
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 \( \lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^+} f(x) = f\left(\frac{1}{2}\right) = 0 \), \( f(x) \) is continuous at \( x = \frac{1}{2} \).
Step 3: Check differentiability at \( x = \frac{1}{2} \).
A function is differentiable at \( x = a \) if the left-hand derivative equals the right-hand derivative at that point.
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 derivatives are not equal at \( x = \frac{1}{2} \). Therefore, \( f(x) \) is not differentiable at \( x = \frac{1}{2} \).