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# Thus we obtain the test if there is to be a maximum

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Thus we obtain the test: if there is to be a maximum or minimum the first derivative which does not vanish must be an even derivative, and there will be a maximum if it is negative, a minimum if it is positive. Examples LVII. 1. Verify the result when φ ( x ) = ( x - a ) m , m being a positive integer, and ξ = a . 2. Test the function ( x - a ) m ( x - b ) n , where m and n are positive integers, for maxima and minima at the points x = a , x = b . Draw graphs of the different possible forms of the curve y = ( x - a ) m ( x - b ) n . 3. Test the functions sin x - x , sin x - x + x 3 6 , sin x - x + x 3 6 - x 5 120 , . . . , cos x - 1, cos x - 1+ x 2 2 , cos x - 1+ x 2 2 - x 4 24 , . . . for maxima or minima at x = 0. 150. B. The calculation of certain limits. Suppose that f ( x ) and φ ( x ) are two functions of x whose derivatives f 0 ( x ) and φ 0 ( x ) are continuous for x = ξ and that f ( ξ ) and φ ( ξ ) are both equal to zero. Then the function ψ ( x ) = f ( x ) ( x ) is not defined when x = ξ . But of course it may well tend to a limit as x ξ . Now f ( x ) = f ( x ) - f ( ξ ) = ( x - ξ ) f 0 ( x 1 ) , where x 1 lies between ξ and x ; and similarly φ ( x ) = ( x - ξ ) φ 0 ( x 2 ), where x 2 also lies between ξ and x . Thus ψ ( x ) = f 0 ( x 1 ) 0 ( x 2 ) . We must now distinguish four cases.

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[VII : 150] ADDITIONAL THEOREMS IN THE CALCULUS 328 (1) If neither f 0 ( ξ ) nor φ 0 ( ξ ) is zero, then f ( x ) ( x ) f 0 ( ξ ) 0 ( ξ ) . (2) If f 0 ( ξ ) = 0, φ 0 ( ξ ) 6 = 0, then f ( x ) ( x ) 0 . (3) If f 0 ( ξ ) 6 = 0, φ 0 ( ξ ) = 0, then f ( x ) ( x ) becomes numerically very large as x ξ : but whether f ( x ) ( x ) tends to or -∞ , or is some- times large and positive and sometimes large and negative, we cannot say, without further information as to the way in which φ 0 ( x ) 0 as x ξ . (4) If f 0 ( ξ ) = 0, φ 0 ( ξ ) = 0, then we can as yet say nothing about the behaviour of f ( x ) ( x ) as x 0. But in either of the last two cases it may happen that f ( x ) and φ ( x ) have continuous second derivatives. And then f ( x ) = f ( x ) - f ( ξ ) - ( x - ξ ) f 0 ( ξ ) = 1 2 ( x - ξ ) 2 f 00 ( x 1 ) , φ ( x ) = φ ( x ) - φ ( ξ ) - ( x - ξ ) φ 0 ( ξ ) = 1 2 ( x - ξ ) 2 φ 00 ( x 2 ) , where again x 1 and x 2 lie between ξ and x ; so that ψ ( x ) = f 00 ( x 1 ) 00 ( x 2 ) . We can now distinguish a variety of cases similar to those considered above. In particular, if neither second derivative vanishes for x = ξ , we have f ( x ) ( x ) f 00 ( ξ ) 00 ( ξ ) . It is obvious that this argument can be repeated indefinitely, and we obtain the following theorem: suppose that f ( x ) and φ ( x ) and their deriva- tives, so far as may be wanted, are continuous for x = ξ . Suppose further that f ( p ) ( x ) and φ ( q ) ( x ) are the first derivatives of f ( x ) and φ ( x ) which do not vanish when x = ξ . Then (1) if p = q , f ( x ) ( x ) f ( p ) ( ξ ) ( p ) ( ξ ) ; (2) if p > q , f ( x ) ( x ) 0 ;
[VII : 151] ADDITIONAL THEOREMS IN THE CALCULUS 329 (3) if p < q , and q - p is even, either f ( x ) ( x ) + or f ( x ) ( x ) → -∞ , the sign being the same as that of f ( p ) ( ξ ) ( q ) ( ξ ) ; (4) if p < q and q - p is odd, either f ( x ) ( x ) + or f ( x ) ( x ) → -∞ , as x ξ + 0 , the sign being the same as that of f ( p ) ( ξ ) ( q ) ( ξ ) , while if x ξ - 0 the sign must be reversed.

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