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By K. F. Riley, M. P. Hobson
This pupil resolution handbook offers entire ideas to the entire odd-numbered difficulties in starting place arithmetic for the actual Sciences. It takes scholars via every one challenge step by step, to allow them to essentially see how the answer is reached, and comprehend any error of their personal operating. scholars will research via instance find out how to arrive on the right solution and enhance their problem-solving talents.
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Additional info for Student Solution Manual for Foundation Mathematics for the Physical Sciences
Example text
15 Show that the lowest value taken by the function 3x 4 + 4x 3 − 12x 2 + 6 is −26. We need to calculate the first and second derivatives of the function in order to establish the positions and natures of its turning points: y(x) = 3x 4 + 4x 3 − 12x 2 + 6, y (x) = 12x 3 + 12x 2 − 24x, y (x) = 36x 2 + 24x − 24. Setting y (x) = 0 gives x(x + 2)(x − 1) = 0 with roots 0, 1 and −2. The corresponding values of y (x) are −24, 36 and 72. Since y(±∞) = ∞, the lowest value of y is that corresponding to the lowest minimum, which can only be at x = 1 or x = −2, as y must be positive at a minimum.
In such a product one of the integers must divide by 3 and at least one of the other integers must be even. Thus each product separately divides by both 3 and 2, and hence by 6, and therefore so does their sum f (x). Thus x being an integer is a sufficient condition for f (x) to be divisible by 6. That it is not a necessary condition can be shown by considering an equation of the form f (x) = x(x + 1)(2x + 1) = 2x 3 + 3x 2 + x = 6m, where m is an integer. As a specific counter-example consider the case m = 4.
Thus the proposal is true for n = N + 1 if it is true for n = N, and this, together with our observation for n = 0, completes the ‘if’ part of the proof. Now suppose that Q(n) = a4 n4 + a3 n3 + a2 n2 + a1 n is divisible by 24 for all integers n ≥ 0. Setting n equal to 1, 2 and 3 in turn, we have a4 + a3 + a2 + a1 = 24p, 16a4 + 8a3 + 4a2 + 2a1 = 24q, 81a4 + 27a3 + 9a2 + 3a1 = 24r, 29 Preliminary algebra for some integers p, q and r. The first of these equations is condition (iii). The other conditions are established by combining the above equations as follows: 14a4 + 6a3 + 2a2 = 24(q − 2p), 78a4 + 24a3 + 6a2 = 24(r − 3p), 36a4 + 6a3 = 24(r − 3p − 3q + 6p), 22a4 − 2a2 = 24(r − 3p − 4q + 8p).