Question

The function \( f(x) = |x - 24| \) is:

• A. Differentiable on \( [0,25] \)
• B. Not continuous at \( x = 24 \)
• C. Neither continuous nor differentiable on \( [0,25] \)
• D. Continuous on \( [0,25] \), but not differentiable on \( [0,25] \)

Hint:

Absolute value functions are continuous everywhere but may have sharp corners where they are not differentiable.

Answer 1

The Correct Option is D

Step 1: Verify Continuity A function is continuous if there are no breaks, jumps, or holes in its graph. The absolute value function \( |x - 24| \) is continuous for all \( x \), hence it is continuous on \( [0,25] \). 
Step 2: Check Differentiability A function is differentiable at a point if the left-hand derivative and right-hand derivative at that point are equal. Since \( f(x) = |x - 24| \), we express it piecewise: \[ f(x) = \begin{cases} x - 24, & \text{if } x \geq 24 \\ -(x - 24) = 24 - x, & \text{if } x<24.\\ \end{cases} \] The left-hand derivative at \( x = 24 \) is: \[ \lim_{h \to 0^-} \frac{f(24 + h) - f(24)}{h} = \lim_{h \to 0^-} \frac{(24 - (24 + h))}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1. \] The right-hand derivative at \( x = 24 \) is: \[ \lim_{h \to 0^+} \frac{f(24 + h) - f(24)}{h} = \lim_{h \to 0^+} \frac{(24 + h) - 24}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1. \] Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = 24 \). 

Final Conclusion: - \( f(x) \) is continuous on \( [0,25] \). - \( f(x) \) is not differentiable at \( x = 24 \). Thus, the correct answer is: 

\[Option (4): \text{Continuous on } [0,25], \text{ but not differentiable at } x = 24.\]