Exercise 6, § 2

On the interval $I = [-1,\ +1]$ consider the vector function $\mathbf{f}$, with values in $\mathbf{R}^2$, defined as follows: $\mathbf{f}(t) = (0,\ 0)$ for $-1 \leq t \leq 0$;

for $0 \leq t \leq 1$. Show that $\mathbf{f}$ is differentiable on $]-1,\ +1[$ but that the image of this interval under $\mathbf{f}'$ is not a connected set in $\mathbf{R}^2$ (cf. exerc. 4 c)).