Question

(a) Let f(x)=√x \) for x ≥ 0. Show f' is unbounded on (0,1] \) but f is nevertheless uniformly continuous on (0,1]. Compare with Theorem.

(b) Show f is uniformly continous on [1,[infinity]).

Answer 1

The function f(x) = sqrt(x) is not uniformly continuous on (0,1], but it is uniformly continuous on [1, ∞). The derivative of f(x) is unbounded on (0,1], but this does not affect the uniform continuity of f(x) on [1, ∞).

Step-by-step explanation:

The function f(x) = sqrt(x) for x ≥ 0 is not uniformly continuous on the interval (0,1]. To show this, we can consider the behavior of the derivative f'(x) = 1/(2*sqrt(x)). As x approaches 0 from the right, f'(x) approaches infinity, which means f' is unbounded on (0,1]. However, despite not being uniformly continuous, the function f(x) = sqrt(x) is uniformly continuous on (0,1] as a consequence of the Heine-Cantor theorem.

On the other hand, the function f(x) = sqrt(x) is uniformly continuous on the interval [1, ∞). This can be shown by proving that for any ε > 0, there exists a δ > 0 such that for any x, y in [1, ∞) with |x - y| < δ, we have |f(x) - f(y)| < ε.