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I indented all code by 4 spaces - even though right below the text box it says code is delimited by 3 backtics. But it still won't let me submit the question!

Here is the full text of the question:

First of all, I'm not a mathematician and this is weird territory for me to venture into. So apologies for the misuse or abuse of any terminology.

I was looking at exponential functions, aka $f(x)=a^x$ with $x\in[0,1]$.

The result of this is an exponential curve in the range of 1 to a.

I wanted to parametrize this with regard to 2 properties - the max value $m$, and the "curviness" $c$.  
I have arrived at the following function:

$$ f(x) = \frac{(c^x)(m-1)}{c-1} + 1, \quad c>0 $$

This function has an obvious problem at $c=1$. But also an obvious (linear) solution:

$$ f(x) = x(m-1) + 1 $$

Now, in Mathematica I got this to work:

    curvePower = 1.1;
    Manipulate[Plot[((curvePower^c)^x-1)(m-1)/(curvePower^c-1)+1,{x, 0,1}, PlotRange->Full], {{c,1,"Curviness"},-100,100, 1},{{m,2,"Max value"},-10,10}]

(a slightly modified form of the function to allow it to bulge both ways - it's amusing animating it and watching it flip around like a jump-rope 😄)

However this does not work:

    expF = ((curvePower^c)^# - 1) (m - 1)/(curvePower^c - 1) + 1 &;
    Manipulate[Plot[expF[x],{x, 0,1}, PlotRange->Full], {{c,1,"Curviness"},-100,100, 1},{{m,2,"Max value"},-10,10}]

Nor is it clear how I can call `expF` by seperately specifying the dependent and independent parameters - without explicitly defining the symbols $c$ and $m$ (which makes the plot in the second example fail to work).

And finally, how can I define the function to actually be

$$ f(x) = 
  \frac{(c^x)(m-1)}{c-1} + 1 & c \neq 1 \\
  x(m-1) + 1 & c = 1

because the top case makes mathematica complain when we set $c=1$.

Other miscellaneous questions:
1. Is that second case what is known as "analytic continuation".
2. Is there a proper term for how extreme the curve is?


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