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Is there a way to write a function so that for any $c$ such that $a < c < b$, $f(c)$ is always the same?

For example, if you had an increasing function up until $0$ at which point the $f(x)$ is $0$ all the way until $10$ when the function starts decreasing again.

I am not looking for horizontal lines or piecewise functions.

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    Why are piecewise functions out of the question? Perhaps smooth bump functions are what you are looking for.2012-09-15

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Although I have no idea why you avoid piecewise defined functions, here is a smooth "flattening" function defined by a differential equation: $ f'(x)=\sqrt{1-f(x)^2},\qquad f(0)=0 $ The solution of the above equation is unique, it agrees with $\sin x$ on $[-\pi/2,\pi/2]$, and agrees with $\mathrm{sign}\, x$ for $|x|>1$. It's just the "sine concatenated with sign" :) but the concatenation comes from the ODE, it's not imposed artificially.

If differential equations are not good either, here is another version: $ g(x)=\min_{0\le t\le |x|} \cos t $ This function agrees with $\cos $ on $[-\pi,\pi]$ and is $-1$ elsewhere. Again, no cases specified in the definition.

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Assuming you are talking about non-piecewise linear functions, you might want to look at absolute values:

$f(x) = a \left | x - p \right | + q$

Where $a$ is the slope, $p$ is the horizontal translation, and $q$ is the vertical translation.

You can also use combinations of absolute values to include more 'sharp'/discontinuous points.


Be aware that absolute values are either defined by piecewise functions:

$\left | x \right | = \begin{cases} -x &\text{if }x<0\\ x&\text{if } x\geq 0\end{cases}$

Or:

$\left | x \right | = \sqrt{x^2}$

As far as I know, there are no exact functions which match your definition, but you can approximate them. However, I would presume that these 'approximation functions' are quite insanely complex.