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.


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


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


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.