Thomas Simpson and Maxima and Minima - Maximizing a polynomial function

To determine the different values of x, when that of 3x⁴ - 28ax³ + 84a² x² - 96a³ x + 48b⁴ becomes a maximum or minimum (Example XXII, page 44).

Let $f(x)$ be the given expression. Then $f'(x) = 12x^3 - 84ax^2 + 168a^2 x - 96a^3$. The critical points of $f(x)$ stem from the equation $12x^3 - 84ax^2 + 168a^2 x - 96a^3 = 0$, which is equivalent to $x^3 - 7ax^2 + 14a^2 x - 8a^3 = 0$. By simple inspection we can realize that the latter equation has solutions $x = a, x = 2a, x = 4a$. Therefore $f'(x) = (x - a)(x - 2a)(x - 4a)$. When $x < a$ we have $x < 2a$ and $x < 4a$, so $f'(x) < 0$; when $x > a$ but $x < 2a$ we get $x < 4a$, thus $f'(x) > 0$. By the first-derivative test we can conclude that $f(x)$ adopts a local minimum at $x = a$. In a similar fashion one can prove that $f(x)$ adopts a local maximum at $x = 2a$ and a local minimum at $x = 4a$.

As a remark (“scholium”) after example XXII (page 45, last paragraph), Simpson discusses the function $g(x) = 24a^3 x - 30a^2 x^2 + 16ax^3 - 3x^4$. We have $g'(x) = 24a^3 - 60a^2 x + 48ax^2 - 12x^3$. Factoring out 12, without much effort we get $g'(x)= -12(x - a)^2 (x - 2a)$. In a small neighborhood of $x = a$ the derivative is positive, so $g(x)$ does not adopt a local maximum or minimum at $x = a$. But it does adopt a local maximum at $x = 2a$ because for $x < 2a$ the derivative is positive, while for $x > 2a$ the derivative is negative.

Remark: Both $f(x)$ and $g(x)$ are easy to analyze through the first-derivative test since their derivatives can be factored without much difficulty. The above-mentioned examples may seem rather ad hoc, nonetheless they illustrate quite well how an important test works.

Editor’s Note: This article was published in 2005.