**To determine the different values of ***x*, when that of 3*x*^{4} - 28*ax*^{3} + 84*a*^{2} *x*^{2} - 96*a*^{3} *x* + 48*b*^{4} becomes a maximum or minimum (Example XXII, page 44).

Let *f*(*x*) be the given expression. Then *f*¢(*x*) = 12*x*^{3} - 84*ax*^{2} + 168*a*^{2} *x* - 96*a*^{3} . The critical points of *f*(*x*) stem from the equation 12*x*^{3} - 84*ax*^{2} + 168*a*^{2} *x* - 96*a*^{3} = 0, which is equivalent to *x*^{3} - 7*ax*^{2} + 14*a*^{2} *x* - 8*a*^{3} = 0. By simple inspection we can realize that the latter equation has solutions *x* = *a*,*x* = 2*a*,*x* = 4*a*. Therefore *f*¢(*x*) = (*x* - *a*)(*x* - 2*a*)(*x* - 4*a*). When *x* < *a* we have *x* < 2*a* and *x* < 4*a*, so *f*¢(*x*) < 0; when *x* > *a* but *x* < 2*a* we get *x* < 4*a*, 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* = 2*a* and a local minimum at *x* = 4*a*.

As a remark ("scholium") after example XXII (page 45, last paragraph), Simpson discusses the function *g*(*x*) = 24*a*^{3} *x* - 30*a*^{2} *x*^{2} + 16*ax*^{3} - 3*x*^{4}. We have *g*¢(*x*) = 24*a*^{3} - 60*a*^{2} *x* + 48*ax*^{2} - 12*x*^{3}. Factoring out 12, without much effort we get *g*¢(*x*)=- 12(*x* - *a*)^{2} (*x* - 2*a*). 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* = 2*a* because for *x* < 2*a* the derivative is positive, while for *x* > 2*a* 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.