Before I ask you my question which I will mark in bold I will tell you what I already gathered so far.

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In a previous result I have showed that the bumpfunction is smooth. The bumpfunction is defined on some intervall $[a,b]$ with

$$\psi(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}}&\mbox{a<x<b}\\0&\mbox{otherwise}\end{cases}$$

For more details I refer to:

[Calculate the n-th derivative for the BUMP function](https://math.stackexchange.com/questions/3113076/calculate-the-n-th-derivative-for-the-bump-function)

I also know that the bump-function can be normalized, i.e 
if $a<b<c<d$ I can construct a smooth function $\psi$ with the properties

$$\phi(x)\begin{cases}=1&\mbox{}b\leq x\leq c\\\in(0,1)&\mbox{if }a<x<b\text{ or }c < x <d\\=0 &\mbox{if }x\leq a \text{ or }x\geq d\end{cases}$$

The function I came up with is:

$$\phi(x)=\psi\_0(x)\begin{cases}\frac{\psi\_1(x)}{\psi\_2(x)+\psi\_1(x)+\psi\_3(x)}&\mbox{a<x<d}\\0&\mbox{otherwise}\end{cases}$$

$$\psi\_1(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{d-x}}&\mbox{a<x<d}\\0&\mbox{otherwise}\end{cases}$$

$$\psi\_2(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}}&\mbox{a<x<b}\\0&\mbox{otherwise}\end{cases}$$

$$\psi\_3(x)\begin{cases}e^{-\frac{1}{x-c}}e^{-\frac{1}{d-x}}&\mbox{c<x<d}\\0&\mbox{otherwise}\end{cases}$$

What I have thought is if we have a value $x$ which is $a<x<b\text{ or }c < x <d$ then the nominator is bigger than the dominator i.e $\psi(x)\in(0,1)$ because all components are also positive, if we have a value which is $c\leq x \leq d$ then $\psi\_2$ and $\psi\_3$ are Zero and the denominator and the Nominator is the same i.e $\psi(x)=1$. Because the function is a product of smooth functions $\psi\_0$ itself is smooth again.

I am however not quite sure about this when I plotted this function on Wolfram Alpha: 

(e^{-1/(x+0.5)}\*e^{-1/(0.5-x)})/((e^{-1/(x+0.5)}\*e^{-1/(-0.25-x)})+(e^{-1/(x+0.5)}\*e^{-1/(0.5-x)})+(e^{-1/(x-0.25)}\*e^{-1/(0.5-x)}))

I got a bump function however the values were not normalized as originally intended.

For $a<b$ there exists also a function $\tau$ for which we have 

$$\tau(x)\begin{cases}=1&\mbox{}x\geq b\\\in(0,1)&\mbox{if }a<x<b\\=0 &\mbox{if }x\leq a \end{cases}$$

Again I first define two helper-functions in this case

$$\tau\_1(x)=\begin{cases} e^{-\frac{1}{x-a}} &\mbox{if } x>a\\0&\mbox{otherwise}\end{cases}$$

$$\tau\_2(x)=\begin{cases} e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}} &\mbox{if } a<x<b\\0&\mbox{otherwise}\end{cases}$$

Then the desired function is 

$$\tau(x)=\frac{\tau\_1(x)}{\tau\_1(x)+\tau\_2(x)}$$

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**Now my question**

Given two smooth functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$

There exist a smooth function $h:\mathbb{R}\rightarrow\mathbb{R}$ for which we have

$$h(x)\begin{cases}=f(x)&\mbox{, if }x\leq a\\=g(x) &\mbox{, if }x\geq b\end{cases}$$

Now if I create a function $\tau$ for the missing interval $(a,b)$ in the Fashion I did above then I have $\lim\_{x\downarrow a}\tau(x)=f(a)$ and $\lim\_{x\uparrow a}\tau(x)=g(b)$ as desired, but the Problem is to Show that it is smooth the function $h(x)$ must also fullfill the condition:

$$\lim\_{x\downarrow a}\tau^{(n)}(x)=f^{(n)}(a)\wedge \lim\_{x\uparrow b}\tau^{(n)}(x)=g^{(n)}(b)\text{ for every n-th derivative} $$

I don't know how I can prove this with the results I already have.

I actually have an idea I take two $\tau$ functions one $\tau\_1$ goes to $0$ as $x$ goes to $b$ and to $1$ as $x$ tends to $a$ and another one with the exact opposite behaviour $\tau\_2$. Then $\tau\_1(x) f(x) + \tau\_2(x) g(x)$ is the function I Need but I Need help to prove that the derivatives of the $\tau$ have the same behaviour on the border. However the Problem is that when applying the productrule to calculate for example the first derivative I would get $\tau\_1'f+\tau\_1f'+\tau\_2'g+\tau\_2g'$. Assuming that the derivatives of $\tau$ have the same convergencebehaviour I would get for $\lim\_{x\uparrow b}h'(x)= g(x)+g'(x)$. But I want only $g'(x)$.

Really hope somebody can help me how I can start 

EDIT:

I am also Looking for a solution that does not involve integrals