# Find the extreme values of the polynomial function

#### lep11

1. Homework Statement
The task is to find the extreme values (and their nature) of the polynomial function . $$f(\vec{x})=x_1x_2+x_1^2+x_2^2+x_3^3+x_4^4.$$

3. The Attempt at a Solution
The critical point is $a=(0,0,0,0)$, which is the solution to $\nabla{f(a)}=0.$ If we form the Hessian matrix $$H_f=\begin{bmatrix} 2& 1 & 0 & 0 \\ 1& 2 & 0 & 0 \\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ ,it's easy to see that $det(H_f)=0$, thus the product of the eigenvalues is zero. This test is inconclusive.

How would one determine the extreme values of $f$ in this case? What is the general approach?

I have also tried

$f(\vec{h})=(h_1+\frac{h_2}{2})^2+(h_4^2)^2+h_3^3,$where the term $h_3^3$
is giving trouble. I cannot factor $f$ such that $f(\vec{h})=(\alpha_1(\vec{h}))^2+...+(\alpha_k(\vec{h}))^2-(\alpha_{k+1}(\vec{h}))^2-...-(\alpha_{k+l}(\vec{h}))^2,$ where $\alpha_i$ are linearly independent linear functions.

Graphical interpretation is not an option either, since we are in $\mathbb{R^4}.$In addition I have tried different values of $x_1,...,x_4$ to try to determine if $a$ is min/max point at all without success.

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#### haruspex

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Maybe I don't understand the question, but what would be the extremal values of x3?

#### lep11

Maybe I don't understand the question, but what would be the extremal values of x3?
Is there something unclear or is the question badly-worded? We are interested in the max/min values of the given function. Since the gradient is zero at (0,0,0,0), something could be happening at that point (local maxima,minima, saddle point). However, the hessian eigenvalues test is inconclusive. I feel like I don't have the tools. Maybe one could determine and show algebrally, but then the right values are needed. So, how would one proceed? How would it help to examine the extremal values of $x^3$?

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#### haruspex

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How would it help to examine the extremal values of x3?
Because you have the term x33, and no other references to x3. So for any given values of the other terms, you can vary x3 freely and see how it affects the result. It is the only such term with an odd power, which makes it rather interesting.

#### lep11

Because you have the term x33, and no other references to x3. So for any given values of the other terms, you can vary x3 freely and see how it affects the result. It is the only such term with an odd power, which makes it rather interesting.
$f(0,0,\frac{1}{n},0)=\frac{1}{n^3}>0$ and $f(0,0,-\frac{1}{n},0)=-\frac{1}{n^3}<0$ for all naturals $n$. Thus $a$ must be saddle point. Correct?

#### haruspex

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$f(0,0,\frac{1}{n},0)=\frac{1}{n^3}>0$ and $f(0,0,-\frac{1}{n},0)=-\frac{1}{n^3}<0$ for all n. Thus $a$ must be saddle point.
Right.

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