. Differential and integral calculus. thera maximum nor a minimum value of y. This fact is also evidentfrom the definitions, since CF is neither greater nor smaller thanthe ordinates which immediately precede and follow it. The illustration further emphasizes the fact, that all the rootsof the equation f\x) = o do not necessarily correspond tomaximum or minimum values oif(x). II. Critical values which renderf(x) = SM (Fig. 16) be the locus of the equation y =/(x).At such points as Z>, F, F, where the tangents are perpen-dicular to the X-axis, we have g =/-(») = ». OA, OB, OC are ther
. Differential and integral calculus. thera maximum nor a minimum value of y. This fact is also evidentfrom the definitions, since CF is neither greater nor smaller thanthe ordinates which immediately precede and follow it. The illustration further emphasizes the fact, that all the rootsof the equation f\x) = o do not necessarily correspond tomaximum or minimum values oif(x). II. Critical values which renderf(x) = SM (Fig. 16) be the locus of the equation y =/(x).At such points as Z>, F, F, where the tangents are perpen-dicular to the X-axis, we have g =/-(») = ». OA, OB, OC are therefore critical values of x. At D,f{x)\0A,= + quantity; at D,f(x)\oA = » ; at D\f\x)\oA — — quantity ; hence at a maximum point f\x)passes through oo from -j- to —. 142 Differential Calculus Similarly, at E,f\x) passes through oo from — to +. At F, /(•*)]0c = °° ; but for values of x a little less and alittle greater than OC we find f\x) = a positive quantity;hence f(x) does not change sign as x passes through the crit-. ical value OC\ hence CF does not represent either a maximumor a minimum value of f(x). This is also evident from thedefinition. It also appears that the roots of the equationf\x) = oo do not necessarily render fix) either a maximum ora minimum. 114. Methods of Investigation for Maximum and MinimumValues. I. By examining the given function. Let/ (x) be the given function, and let a be any one of thecritical values found by equating fr(x) to o or oo , or both. Then, by the definition, § in, we have, h being a very smallquantity, (/(«)>/(«-*)? }f(a)>/(a + A) f(a)<f{a-h)f(a)<f{a + h) For a maximum For a minimum Thus, let then, 0) (2) /<*) 3 a* 6x + 8. Maxima and Minima 143 As no finite value of x will render f(x) = oo , we equate itto zero; hence x2 — 6 x + 8 = (x — 2) (x — 4) = o,.•. x = 2 and a: = 4, are the critical values of x. Substituting values a little less and Xsa little greater than x = 2 in the given function, 3 .x
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