Finding the equation of a rational function

Question

I have a formula for a rational fuction in the form $$\frac{1}{(a(x-b)^c)}=y$$ and I am given the points $(0,30.8493),(0.75,1.1392)$, and $(1,0.2838)$. When simplifying I always seem to reach a dead end with a no solution answer. I know that there is atleast one solution because I was able to find a very close graphical solution to the problem by manimulating $a, b$, and $c$ until it worked. With my current skill level in math, I am unable to figure out this problem.

Answer 1

Using Matti P.'s comment, use the equations $$a\,b^c = \frac{1}{y_1}\tag 1$$ $$a \,\left(\frac{3}{4}-b\right)^c= \frac{1}{y_2}\tag 2$$ $$a \,\left(1-b\right)^c= \frac{1}{y_3}\tag 3$$

Use $(1)$ to eliminate $a$ $$a=\frac{b^{-c}}{y_1}$$ Replace in $(3)$ to get $$\frac{(1-b)^c b^{-c}}{y_1}=\frac{1}{y_3} \implies c=\frac{\log \left(\frac{y_1}{y_3}\right)}{\log (1-b)-\log (b)}$$

Replace in $(2)$ to get an equation in $b$. Plot it for $0 < b < 1$ and zoom more and more to "see" a solution close to $b\approx0.215$ that you could refine using Newton method (use numerical derivatives); it would coverge very fast.

When done, go back to $a$ and $c$ to get the final results.

Edit (for the fun of it)

If you work carefully, the last equation in $b$ can rewrite $$\frac{\tanh ^{-1}\left(1-\frac{8}{3}b\right)}{\tanh ^{-1}(1-2 b)}=\frac{\log \left(\frac{y_1}{y_2}\right)}{\log \left(\frac{y_1}{y_3}\right)}$$ Since $y_1 > y_2 > y_3$, the rhs is $<1$ making the solution to be in $\left(0,\frac 12\right)$.

To make life easier, we can approximate the lhs building its simplest Padé approximant at $b=\frac 14$ that is to say $$\frac{\tanh ^{-1}\left(1-\frac{8}{3}b\right)}{\tanh ^{-1}(1-2 b)}\sim \frac{\alpha_0+\alpha_1\left(b-\frac{1}{4}\right)} {1+ \alpha_2\left(b-\frac{1}{4}\right)}$$ where $$\alpha_0=\frac{\tanh ^{-1}\left(\frac{1}{3}\right)}{\tanh ^{-1}\left(\frac{1}{2}\right)}\approx 0.63093$$ $$\alpha_1=-\frac{2 \left(4 \log (2) (\log (4)-9)-27 \coth ^{-1}(2) \left(\coth ^{-1}(3)-3\right)\right)}{3 \log \left(\frac{19683}{256}\right) \coth ^{-1}(2)}\approx -5.09907$$ $$\alpha_2=-\frac{2 \left(8 \log (2) (\log (3)-4)+9 \left(8-3 \coth ^{-1}(2)\right) \coth ^{-1}(2)\right)}{3 \log \left(\frac{19683}{256}\right) \coth ^{-1}(2)}\approx -4.28028$$ Using your numbers, this would give $b\approx 0.215200$ which is almost the exact solution.