I am searching for a function $f\colon (1,2]\to \mathbb{R}$ that fullfills
- $f(x)\to \infty$ as $x\to 1$
- $f(2)=1,\quad f(3/2)=1+\sqrt{2},\quad f(4/3)=3$
I noticed that $g(x)=\frac{1}{x-1}$ nearly fullfills this except that $g(3/2)=2$. Is there maybe a way to slightly change the function so that it works? (Btw it would be nice if the function would be smooth and has a most simple form possible).
Start from
$f(x)=\dfrac{a x^2+b x+c}{x-1}$
$f(2)=4 a + 2 b + c = 1$
$f(3/2)=2 \left(\frac{9 a}{4}+\frac{3 b}{2}+c\right)=\sqrt{2}+1$
$f(4/3)=3 \left(\frac{16 a}{9}+\frac{4 b}{3}+c\right)=3$
Solve the system and get
$a= -6 \left(\sqrt{2}-1\right),b= 20 \left(\sqrt{2}-1\right),c= 17-16 \sqrt{2}$
Plug into the function
$$f(x)=\frac{-6 \left(\sqrt{2}-1\right) x^2+20 \left(\sqrt{2}-1\right) x-16 \sqrt{2}+17}{x-1}$$
In the picture below you see the graph of the function on $(1,2]$ and the plot of the given points $(2,1);\;(3/2,2);\;(4/3,3)$
Hope this is useful
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