f (c) < (f (x) + ε) ≤ (k + ε) ——– (1) Similarly, values of x between c and c + δ that are not contained in A, such that. 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 /LastChar 255 /Ascent 750 >> 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 Therefore by the definition of limits we have that ∀ M ∃ K s.t. 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 0 0 0 339.29] 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 /Length 3528 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 /FontBBox [-119 -350 1308 850] /Name /F2 (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. 21 0 obj << /FontFile 17 0 R 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 /FontFile 8 0 R /Ascent 750 >> /FontBBox [-100 -350 1100 850] Typically, it is proved in a course on real analysis. /FirstChar 33 Since both of these one-sided limits are equal, they must also both equal zero. 27 0 obj /FontDescriptor 15 0 R /FirstChar 33 /Name /F1 endobj 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 /FontName /YNIUZO+CMR7 First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). /FontFile 26 0 R 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 0 693.75 954.37 868.93 797.62 844.5 935.62 886.31 677.58 769.84 716.89 880.04 742.68 /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring /CapHeight 683.33 /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 The proof that $f$ attains its minimum on the same interval is argued similarly. Examples 7.4 – The Extreme Value Theorem and Optimization 1. In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. /BaseFont /NRFPYP+CMBX12 endobj 9 0 obj The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 892.86 892.86 892.86 1138.89 1138.89 892.86 892.86 1138.89 0 0 0 0 0 0 0 0 0 0 0 /XHeight 430.6 Suppose that is defined on the open interval and that has an absolute max at . (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. endobj 769.85 769.85 769.85 769.85 708.34 708.34 523.81 523.81 523.81 523.81 585.32 585.32 /LastChar 255 772.4 639.7 565.63 517.73 444.44 405.9 437.5 496.53 469.44 353.94 576.16 583.34 602.55 State where those values occur. /LastChar 255 endobj 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 /FontFile 11 0 R Depending on the setting, it might be needed to decide the existence of, and if they exist then compute, the largest and smallest (extreme) values of a given function. /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 0 0 0 0 0 0 575] /XHeight 430.6 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi Note that $g(x) \gt 0$ for every $x$ in $[a,b]$ and $g$ is continuous on $[a,b]$, and thus also bounded on this interval (again by the Boundedness theorem). 493.98 437.5 570.03 517.02 571.41 437.15 540.28 595.83 625.69 651.39 0 0 0 0 0 0 If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. endobj /Type /Font 18 0 obj 30 0 obj 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 /LastChar 255 /Type /Font 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef >> /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 /Name /F6 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. By the Extreme Value Theorem there must exist a value in that is a maximum. Also we can see that lim x → ± ∞ f (x) = ∞. /Subtype /Type1 We can choose the value to be that maximum. 13 0 obj 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Since f never attains the value M, g is continuous, and is therefore itself bounded. Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 /FontName /IXTMEL+CMMI7 767.86 876.99 829.37 630.96 815.48 843.26 843.26 1150.8 843.26 843.26 692.46 323.41 /Subtype /Type1 The rest of the proof of this case is similar to case 2. 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 /ItalicAngle -14 /FontBBox [-116 -350 1278 850] Given that $g$ is bounded on $[a,b]$, there must exist some $K \gt 0$ such that $g(x) \le K$ for every $x$ in $[a,b]$. Proof of the extreme value theorem By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) M of f exists. 769.85 769.85 892.86 892.86 523.81 523.81 523.81 708.34 892.86 892.86 892.86 0 892.86 /FontFile 20 0 R Theorem 1.1. Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). 519.84 668.05 592.71 661.99 526.84 632.94 686.91 713.79 755.96 0 0 0 0 0 0 0 0 0 312.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave >> 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 /Type /FontDescriptor Thus for all in . /Subtype /Type1 Extreme Value Theorem: If a function f (x) is continuous in a closed interval [a, b], with the maximum of f at x = c 1 and the minimum of f at x = c 2, then c 1 and c 2 are critical values of f. Proof: The proof follows from Fermat’s theorem and is left as an exercise for the student. /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 625 500 625 513.31 343.75 562.5 625 312.5 343.75 593.75 312.5 937.5 625 562.5 625 We prove the case that $f$ attains its maximum value on $[a,b]$. 24 0 obj /Ascent 750 /FontFile 23 0 R For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. /Type /Font << /Type /FontDescriptor 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 It is necessary to find a point d in [ a , b ] such that M = f ( d ). >> /Subtype /Type1 /Flags 4 The extreme value theorem: Any continuous function on a compact set achieves a maximum and minimum value, and does so at specific points in the set. We needed the Extreme Value Theorem to prove Rolle’s Theorem. << /Descent -951.43 endobj 19 0 obj >> Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. Both proofs involved what is known today as the Bolzano–Weierstrass theorem. 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 xڵZI����WT|��%R��$@��������郦J�-���)�f��|�F�Zj�s��&������VI�$����m���7ߧ�4��Y�?����I���ԭ���s��Css�Өy������sŅ�>v�j'��:*�G�f��s�@����?V���RjUąŕ����g���|�C��^����Cq̛G����"N��l$��ӯ]�9��no�ɢ�����F�QW�߱9R����uWC'��ToU���
W���sl��w��3I�뛻���ݔ�T�E���p��!�|�dLn�ue���֝v��zG�䃸� ���)�+�tlZ�S�Q���Q7ݕs�s���~�����s,=�3>�C&�m:a�W�h��*6�s�K��C��r��S�;���"��F/�A��F��kiy��q�c|s��"��>��,p�g��b�s�+P{�\v~Ξ2>7��u�SW�1h����Y�' _�O���azx\1w��%K��}�[&F�,pЈ�h�%"bU�o�n��M���D���mٶoo^�� *`��-V�+�A������v�jv��8�Wka&�Q. /Name /F5 It tends to zero in the limit, so we exploit that in this proof to show the Fundamental Theorem of Calculus Part 2 is true. /FontDescriptor 21 0 R /LastChar 255 endobj 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Descent -250 /StemV 80 Sketch of Proof. Consider the function g = 1/ (f - M). << We show that, when the buyer’s values are independently distributed /FontBBox [-115 -350 1266 850] 12 0 obj which implies (upon multiplication of both sides by the positive $M-f(x)$, followed by a division on both sides by $K$) that for every $x$ in $[a,b]$: This is impossible, however, as it contradicts the assumption that $M$ was the least upper bound! 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