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I know that if you have a bump function $B$

$$B(x) = \begin{cases} e^\frac{-1}{x}, & x>0 \\[2ex] 0, & x\le0 \end{cases}$$

and some other arbitrary smooth function $f$ (i.e. $f(x)=x^2$) and create a secondary function to bind both $B$ and $f$ to make $f=B$ for some interval and $f\neq B$ anywhere outside of that interval (as done in the comments here).

However, if you have two functions, $f$ and $g$ (say $g(x)=x$), is there a way to use a bump function to smoothly "connect" these two functions (you probably need to make sure there is some sort of intersection between $f$ and $g$ anyways, $g(x)=x$ may be a poor example).

The behavior I am looking for is that, when $B\le0$, $B=f$ and when $B\ge1$, $B=g$, and across the interval $(0,1)$ $B$ is a $C^\infty$ function where $B\neq f$ and $B\neq g$. I'm sure there is some sort of restriction between what $f$ and $g$ can be, which may not be properly represented in this example.

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The function that's more useful for your purpose is the so-called bump function:

$$f : \mathbb{R} \to \mathbb{R}, x \mapsto \begin{cases} \exp\left( -\frac{1}{1 - x^2} \right) & \text{ if } |x| < 1, \\ 0 & \text{ else}. \end{cases}$$

The reason this one is nicer is because it is also a smooth function, but this one is compactly supported, so outside of the small region $(-1,1)$, this is zero, so it's a useful tool for making tiny fixes to your function around small areas. It basically behaves the same way that your function $e^{-\tfrac{1}{x}}$ behaves at zero, but at two different points.

What's more, consider what happens if you integrate this function (and for reasons you'll see in a moment, we also normalize the integral

$$F : \mathbb{R} \to \mathbb{R}, x \mapsto \frac{\int_{-\infty}^x f(y) dy }{\int_{-\infty}^\infty f(y) dy}.$$

This is clearly again a smooth function, and it has the property that if $x < -1$, then $F(x) = 0$, and if $x > 1$, then $F(x) = 1$. Hence, if you have any other function $h : \mathbb{R} \to \mathbb{R}$, the function $F \cdot h$ will have

$$(F \cdot h )|_{(-\infty,-1)} = 0, \quad (F \cdot h )|_{(1,\infty)} = h|_{(1,\infty)}. $$

Hence, if you want to glue two functions $g$ and $h$ together, multiply one of them with $F$, multiply the other one with a horizontally flipped $F$, and add them together! (of course, this $F$ does the connecting across the interval $(-1,1)$ and not $(0,1)$ as you asked, but that's just a matter of translating and rescaling the function).

If you would instead like to stay with your $e^{-1/x}$-function, you can also look into the answers to this question, there are some constructions which are even more elementary :)

Note in particular: you do not need any specific requirements for your functions, you can glue all two functions together that you like!

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