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I would like to plot some graphs with the general form $f(x)=ax^3+bx^2+cx+d$ on the domain $[0.5,1]$.

The parameters $(a,b,c,d)$ are dependent on some other variables $(p,q,r)$ and they satisfy all of the followings:

  1. $f'(0.5)=0.75a+b+c=p$
  2. $f'(1)=3a+2b+c=q$
  3. $f(1)=a+b+c+d=r$
  4. $f(0.5)=0.125a+0.25b+0.5c+d=0.5$

where all $p$ and $q$ are positive real numbers and $r$ is in $(0.5,1)$.

I solved the above system of equations and managed to obtain a set of solution as follows:

$a=4p+4q-16r+8$

$b=-10p-8q+36r-18$

$c=8p+5q-24r+12s$

$d=-2p-q+5r-2$

BUT I am wondering if there IS actually another set of solutions which also satisfies the above conditions. The solution above gave me some plots where $f''(x)$ is not strictly positive (i.e. $6ax+2b>0$) and this is not what I want. I need a solution which gives me curves whose slopes are always increasing, if not constant, but I do not know how to do it.

Thanks in advance!

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EDIT:

I am actually seeking a function, not necessarily the cubic function above, that can satisfy (1)-(4) above, as will as $f''(x)$ is positive and increasing (i.e. $f'''(x)$, given the ranges of $(p,q,r)$ above.

So whether to get an alternative solution for the above system is not really the ultimate concern...

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  • $\begingroup$ There seem to be two related but different questions here. The system (1) - (4) you have formulated indeed has a unique solution. However, it does not reflect the subsequent statements that the slopes are increasing on $[1/2, 1]$. Which of these is the question you really need to address? If it's the latter, do you really mean that the slopes increase (that is, get larger with larger $x$) or do you actually mean that they should be nonnegative? $\endgroup$
    – whuber ♦
    Commented Dec 4, 2017 at 20:31
  • $\begingroup$ Actually, I am finding an equation, not necessarily $f(x)=ax^3+bx^2+cx+d$, which I can map $(p,q,r)$ on and has to satisfy (1)-(4), $f''(x)$ positive (yes, the slopes should increase with x), AND something that I forgot to mention: $f'''(x)$ positive. I tried $f(x)=ax^2+bx+c$ before, but this does not generate a fitting that I want, because I want the increase in slope to be mild at first and more vigorous later (i.e. $f''(x)$ will increase with x). So I came to this cubic equation, which did not give me what I want because now $f''(x)$ is not always positive. $\endgroup$
    – Matthew Hui
    Commented Dec 5, 2017 at 3:01
  • $\begingroup$ I also tried some exponential function but I have not been able to solve them... $\endgroup$
    – Matthew Hui
    Commented Dec 5, 2017 at 3:02

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