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Hi everyone! I have a question about how to find the closed form of a function defined by

$$\phi(\theta)=\inf_{x\geq 2}f(x;\theta)\equiv\inf_{x\geq 2}\frac{(x+2)^2}{\frac{1}{\theta}\left(\frac{x-1}{2}-\frac{1}{x}\right)+\frac{x^2-1}{16}},\ \ \theta>0$$

Since finding the minimum of $f(x;\theta)$ leads to solving a cubic polynomial. Could someone help me to characterize $\phi(\theta)$ please? Many thanks!

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1 Answer 1

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If $\theta>2$ then $\phi(\theta)=16$. For $\theta<2$, let $$t(x)=\frac{4(x^3-4x^2-6x-4)}{x^2(1+2x)},$$ $$g(x)=f(x,t(x)).$$ The function t is invertible from $[x_m,\infty)$ to $[0,2)$, where $x_m$ is the root of t(x)=0, approximately 5.28. We have $\phi(\theta)=g(t^{-1}(\theta))$.

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