Let be $\displaystyle f(x):=g(1-g(x)); x \in \mathbb{R}$, where $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}\text{; } x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} & \text{if }x>0\\0 & \text{if }x\leq 0\end{cases}$

Let be $\displaystyle a, b \in \mathbb{R}$ and $\displaystyle h(x)=f(a-x)f(x-b)$.

Prove that: $\displaystyle h \in C^\infty;\: h\geq0; h(x)=1 \text{ if } x\in [a,b] \text{ and } h(x)=0 \text{ if } x \not\in ]a-1,b+1[ $.

I know that $\displaystyle g \in C^\infty$ but not analyitic in 0. I also know that $\displaystyle f$ is a monotonically decreasing function $\displaystyle f(x)=1 \text{ if } x \leq 0 \text{ and } f(x)=0 \text{ if } x>0$.

Thank you very much in advance!