Step 1: Understanding the Concept:
The function \(f(x)\) is the area under the greatest integer function \([y]\).
Integrals of piecewise constant functions (like the floor function) are always continuous.
Differentiability is determined by the Fundamental Theorem of Calculus: if \(f(x) = \int_0^x g(y) dy\), then \(f'(x) = g(x)\) at points where \(g\) is continuous.
Step 3: Detailed Explanation:
1. Continuity:
Let \(x = n + f\), where \(n \in \mathbb{I}\) and \(0 \le f<1\).
\(f(x) = \int_0^1 0 dy + \int_1^2 1 dy + ....... + \int_{n-1}^n (n-1) dy + \int_n^x n dy\).
\(f(x) = [0 + 1 + 2 + ....... + (n-1)] + n(x - n) = \frac{n(n-1)}{2} + n(x - n)\).
At any integer point \(x = k\), the limit from the left and right exists and equals \(\frac{k(k-1)}{2}\). Thus, \(f(x)\) is continuous for all \(x \ge 0\).
2. Differentiability:
Using the Fundamental Theorem of Calculus, \(f'(x) = [x]\) at non-integer points.
At integer points \(x = k\):
Right-hand derivative: \(\lim_{h \to 0^+} \frac{f(k+h) - f(k)}{h} = \lim_{h \to 0^+} \frac{[k \cdot h]}{h} = k\).
Left-hand derivative: \(\lim_{h \to 0^-} \frac{f(k+h) - f(k)}{h} = \lim_{h \to 0^-} \frac{[(k-1) \cdot h]}{h} = k - 1\).
Since \(LHD \neq RHD\) at integer points, \(f\) is not differentiable there.
Step 4: Final Answer:
\(f\) is continuous everywhere but differentiable except at integer points.