The function \( f(x) = |x - 24| \) is considered. We need to determine its continuity and differentiability on the interval \([0, 25]\).
Step 1: Analyze Continuity
The absolute value function \( |x| \) is continuous everywhere, meaning \( f(x) = |x - 24| \) is also continuous everywhere, including on \([0, 25]\).
Conclusion on Continuity: The function \( f(x) \) is continuous on \([0, 25]\).
Step 2: Analyze Differentiability
The differentiability of \( f(x) = |x - 24| \) requires us to examine the point where \( x = 24 \) because absolute value functions may not be differentiable at points where the expression inside the absolute value changes sign.
Specifically, at \( x = 24 \), the left-hand derivative and right-hand derivative are calculated as follows:
At \( x = 24 \), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\). Since these derivatives are not equal, \( f(x) \) is not differentiable at \( x = 24 \).
Conclusion on Differentiability: The function is not differentiable at \( x = 24 \), but differentiable elsewhere on \([0, 25]\).
Therefore, the function \( f(x) = |x-24| \) is continuous on \([0,25]\), but not differentiable on \([0,25]\), particularly due to the point \( x = 24 \).
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:
\[Option (4): \text{Continuous on } [0,25], \text{ but not differentiable at } x = 24.\]