Partial solution: construct 2 functions g(x) and h(x), to satisfy:
(looks hard to do but isn't, the RHS is just a number)
Now, consider:
as required
as required
as required
as required
i cant guarantee f(x) will be monotonic though
Hello there - bit of an unusual question and I'm not quite sure if it belongs in this Calculus section or not. It sorta involves derivatives so here it goes:
I have been looking for monotonic equations that pass through (0,0) and (1,1) with various slopes at x=0 and x=1. For example, I already have:
CLASS 1,2: f(x)=x^n
when n>1: concave up, f'(0) = 0, f'(1) = n
when 0<n<1: concave down, f'(0) = inf, f'(1) = n
CLASS 3,4: f(x) = 1-(1-x)^n
when n>1: concave down, f'(0) = n, f'(1) = 0
when 0<n<1: concave up, f'(0) = n, f'(1) = inf
CLASS 5,6: f(x) = (1-(1-x)^n)^(1/n)
when n>1: concave down, f'(0) = inf, f'(1) = 0
when 0<n<1: concave up, f'(0) = 0, f'(1) = inf
What I am looking for are the equations that will have:
f'(0) = m
f'(1) = n
Thus, 2 different finite slopes at x=0 and x=1 allowing both concave up and down curves
Thank you for any help. I have asked math professors at my college but still no answer. Sorry for the long post...
Terry
Partial solution: construct 2 functions g(x) and h(x), to satisfy:
(looks hard to do but isn't, the RHS is just a number)
Now, consider:
as required
as required
as required
as required
i cant guarantee f(x) will be monotonic though
Thanks for the help! With the above hint as a starting point, I was able to figure out the following general equation for curves with slope 1/n at (0,0) and slope n at (1,1):
y= (x^n^(1/n)) * (x^n) + (1-x^n^(1/n)) * (1-(1-x)^(1/n))
By changing the 'coefficients' from x^2 and (1-x^2) to (x^n^(1/n)) and (1-x^n^(1/n)), I found that the integrals of the curve on both sides of the line y-x+1 become equal although, I am still not convinced that the curve is symmetrical over that line...