One thing might do is find a function $f$ whose graph is qualitatively the same as yours on the "left part", that is over $[0,C/2]$, say, and relatively constant on the right part, $[C/2, C]$.
Then find a function $g$ whose graph matches yours on the right part and is relatively constant on the left part.
Taking the sum $f+g$, you'll obtain a function whose graph is qualitatively the same as yours. The fact that $f$ and $g$ are nearly constant where they are is important, as this will insure that the sum $f+g$ will still have more or less the right shape.
However, the sum $f+g$ may not be 0 at $x=C$; you may need to vertically shift the function by adding a constant.
Here to get the bell shaped part on the left, you might use $f(x)=\exp(-x^2)$. For the shape of the right, the function $g(x)=.2|C-x|^{1/3}$ seems to work nicely.
So take $h(x)= \exp(-x^2) + .2|C-x|^{1/3}$. Even better would be $h(x)= \exp(-x^2) + .2|C-x|^{1/3} -\exp(-C^2)$, as this will insure that $h(C)=0$.
Well, that's almost right, except $h(0)$ is not $1$. But to ameliorate this, we can vertically scale, take $h(x)=a\bigl( \exp(-x^2) + .2|C-x|^{1/3} -\exp(-C^2)\bigr)$ where $a$ is chosen so that $h(0)=1$.
Actually, my choices of $f$ and $g$ are somewhat unsatisfactory (in particular, the function $g$ isn't nearly constant on the left part; perhaps the choice I made in my comment to your question ($g(x) = -{1\over C-x}$) would be a better choice). But I hope the method described is of help.