**To determine the different values of x, when that of 3x^{4} - 28ax^{3} + 84a^{2} x^{2} - 96a^{3} x + 48b^{4} 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.*