This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1889 edition. Excerpt: ... /(a-7l)-/(-)=/--/'-+/"(a)4--, (1) and /(o+A)-/(a)=/"(a)+/"'(a)+/CT(o)'+-(2) If h be taken very small, the sign of the second member of either (1) or (2) will be the same as the sign of its first term. Hence, if f"a) is negative, f(a) is greater than both f(a--h) and f(a + h), and therefore a maximum; ...
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1889 edition. Excerpt: ... /(a-7l)-/(-)=/--/'-+/"(a)4--, (1) and /(o+A)-/(a)=/"(a)+/"'(a)+/CT(o)'+-(2) If h be taken very small, the sign of the second member of either (1) or (2) will be the same as the sign of its first term. Hence, if f"a) is negative, f(a) is greater than both f(a--h) and f(a + h), and therefore a maximum; while, if f"(a) is positive, f(a) is less than both f(a--h) and /(a + h), and therefore a minimum. If f"(a) is 0, and /"'(a) is not 0, /(a) is neither greater than both f(a--h) and f(a + h), nor less than both, and is therefore neither a maximum nor a minimum. If/"'(a), as well as f"(a), is 0, and /IT(-) is negative, f(a) is greater than both /(a--h) and f(a + h), and therefore a maximum; while, if /, T(-) is positive, fa) is a minimum, and so on. gS 7a;4-6 Hence, if a is a critical value obtained from f'(x) = 0, substitute a for x in the successive derivatives of f (x). If the first derivative that does not reduce to 0 is of an odd order, f(a) is neither a maximum nor a minimum; but, if the first derivative that does not reduce to 0 is of an even order, f(a) is a maximum or a minimum, according as this derivative is negative or positive. 113. Maxima and minima occur alternately. Suppose that a 6, and that /(a) and /(6) are maxima of f(x). When x--a + h, fx) is decreasing; and, when x=b--h, f(x) is increasing, h being very small. But, in passing from a decreasing to an increasing state, f(x) must pass through a minimum. Hence, between two maxima, there must be at least one minimum. In like manner, it can be proved that between two minima there must be at least one maximum. 114. The solution of problems in maxima and minima is sometimes facilitated by the following...
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