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For any fix $t\in\mathbb{R}$, show that the function $F_{t}(y)=y^4+ty^2+t^2y$ achieves its global minimum at a single point $y_{0}(t)$. I have made some partial attempts to solve this question, but I am stuck at one of the steps and would like to ask for hints.

$\textbf{My attempt}$: Fix any $t\in\mathbb{R}$. Since $\lim_{y\rightarrow\infty}F_{t}(y)=\lim_{y\rightarrow -\infty}F_{t}(y)=\infty$, $F_t(y)$ achieves its global minimum at a point $y_{0}$. We now show that there can only be one values of $y$ where $F_{t}(y)$ achieves its global minimum.

Case 1: If $t=0$ then it is quite obvious.

Case 2: Assume $t>0$, then $\frac{dF_{t}(y)}{dy}=4y^3+2ty+t^2$, and $\frac{d^2F_{t}(y)}{dy^2}=12y^2+2t$. Since the second derivative of $F_{t}(y)$ is always positive, the first derivative of $F_{t}(y)$ is always strictly increasing, and hence must be injective. Furthermore, $\lim_{y\rightarrow -\infty}\frac{dF_{t}(y)}{dy}=-\infty$, and $\lim_{y\rightarrow\infty}\frac{dF_{t}(y)}{dy}=\infty$. By continuity and previous argument, the first derivative of $F_{t}(y)$ has exactly one zero. Therefore, $F_{t}(y)$ achieves its global minimum only at a single point $y_{0}(t)$.

Case 3: t<0. I am stuck at this step. A hint would be useful.

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This function can only achieve global minima at two distinct points when it is of the form

$$((x-a)(x-b))^2+c.$$

As there is no cubic term, $a=-b$ and the equation is in fact

$$(x^2-a)^2+c=x^4-2ax^2+a^2+c.$$

As there is no linear term, we must have $t=0$, which also imposes $a=0$.

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  • $\begingroup$ the first sentence that you wrote is really helpful. $\endgroup$ – KnobbyWan Oct 2 '18 at 13:45

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