Page:Calculus Made Easy.pdf/118

The second remark is one which (if you have any wits of your own) you will probably have already made: namely, that this much-belauded process of equating to zero entirely fails to tell you whether the $x$ that you thereby find is going to give you a maximum value of $y$ or a minimum value of $y$. Quite so. It does not of itself discriminate; it finds for you the right value of $x$ but leaves you to find out for yourselves whether the corresponding $y$ is a maximum or a minimum. Of course, if you have plotted the curve, you know already which it will be.

For instance, take the equation:

$y=4x+{\frac {1}{x}}$.

Without stopping to think what curve it corresponds to, differentiate it, and equate to zero:

${\frac {dy}{dx}}=4-x^{-2}=4-{\frac {1}{x^{2}}}=0$;

whence

$x={\tfrac {1}{2}}$;

and, inserting this value,

$y=4$

will be either a maximum or else a minimum. But which? You will hereafter be told a way, depending upon a second differentiation, (see Chap. XII. p. 112). But at present it is enough if you will simply try any other value of $x$ differing a little from the one found, and see whether with this altered value the corresponding value of $y$ is less or greater than that already found.