If $f'$ exists on an open interval, and there is some point $c$ where $f'(c) > 0$, then there exists a d-neighborhood $\{x \in \mathbb{R} : |x - c| < d\} = V_d(c)$ around c in which $f'(x) > 0$ for all $x \in V_d(c).$

1.How to presage this is false?

2.Where did this counterexample feyly emanate from?

f is differentiable everywhere, including $x = 0$.

To prove $f'(x)$ takes on negative values..., compute $f'(x) = \dfrac{1}{2} + 2x\sin\dfrac{1}{x} - \cos \dfrac{1}{x}$.

3.How does $x(k)$ 'enter'?

4.Why will $g'(x_k) \in (-d, d)$ for k large enough?