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If f is a real valued differentiable function satisfying |f(x) - f(y)| ≤ (x - y)², x, y ∈ R and f(0) = 0, then f(1) equals
- -1
- 0
- 2
- 1
Correct answer: 0
Solution
Dividing by |x-y| gives |f(x)-f(y)|/|x-y| <= |x-y| -> 0, so f'(x)=0 for all x and f is constant. With f(0)=0, f(1)=0.
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