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 /Type /FontDescriptor /CapHeight 686.11 593.75 459.49 443.75 437.5 625 593.75 812.5 593.75 593.75 500 562.5 1125 562.5 562.5 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. Also we can see that lim x → ± ∞ f (x) = ∞. The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. Thus for all in . /Flags 4 << /CapHeight 683.33 >> /FontBBox [-119 -350 1308 850] /FontDescriptor 27 0 R /BaseFont /NRFPYP+CMBX12 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 /FontName /IXTMEL+CMMI7 << /BaseEncoding /WinAnsiEncoding /Name /F4 569.45] /XHeight 430.6 >> It may then be shown that: f 0 (c) = lim h → 0 f (c + h)-f (c) h = 0, using that fact that if f (c) is an absolute extremum, then f (c + h)-f (c) h is both ≤ 0 and ≥ 0. << 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 Both proofs involved what is known today as the Bolzano–Weierstrass theorem. /FirstChar 33 /Descent -250 /StemV 80 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 Letfi =supA. << /Flags 68 /FirstChar 33 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 27 0 obj /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 777.78 625 916.67 750 777.78 680.56 777.78 736.11 555.56 722.22 750 750 1027.78 750 This theorem is sometimes also called the Weierstrass extreme value theorem. /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 By the Extreme Value Theorem there must exist a value in that is a maximum. << /Filter [/FlateDecode] It states the following: If a function f (x) is continuous on a closed interval [ a, b ], then f (x) has both a maximum and minimum value on [ a, b ]. 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 /Descent -250 /Subtype /Type1 19 0 obj /FontFile 20 0 R f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. Then $f(x) \lt M$ for all $x$ in $[a,b]$. endobj endobj First we will show that there must be a finite maximum value for f (this >> State where those values occur. 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 /ItalicAngle 0 /LastChar 255 Examples 7.4 – The Extreme Value Theorem and Optimization 1. 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. /LastChar 255 /Type /FontDescriptor /BaseFont /TFBPDM+CMSY7 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 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 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 Because f(x) is continuous on [a, b], by the Extreme Value Theorem, we know that f(x) will have a minimum somewhere on [a, b]. 0 0 0 339.29] /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 /BaseFont /IXTMEL+CMMI7 /Type /Font /Ascent 750 Proof of Fermat’s Theorem. The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 /LastChar 255 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500] 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 We will first show that \(f\) attains its maximum. /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). endobj /LastChar 255 Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. Sketch of Proof. /Ascent 750 /Flags 68 /FontFile 23 0 R Therefore by the definition of limits we have that ∀ M ∃ K s.t. Thus, before we set off to find an absolute extremum on some interval, make sure that the function is continuous on that interval, otherwise we may be hunting for something that does not exist. This makes sense because the function must go up (as) and come back down to where it started (as). /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] /FontBBox [-100 -350 1100 850] /Type /FontDescriptor The standard proof of the first proceeds by noting that f is the continuous image of a compact set on the interval [a,b], so it must itself be compact. 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 10 0 obj /BaseFont /PJRARN+CMMI10 /FontName /NRFPYP+CMBX12 Theorem 6 (Extreme Value Theorem) Suppose a < b. 819.39 934.07 838.69 724.51 889.43 935.62 506.3 632.04 959.93 783.74 1089.39 904.87 /Ascent 750 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 Suppose there is no such $c$. If f : [a;b] !R, then there are c;d 2[a;b] such that f(c) •f(x) •f(d) for all x2[a;b]. As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. /Name /F6 Typically, it is proved in a course on real analysis. >> /XHeight 444.4 https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem >> /FirstChar 33 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. /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 /Encoding 7 0 R /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 18 0 obj In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. We look at the proof for the upper bound and the maximum of f. 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. The proof that $f$ attains its minimum on the same interval is argued similarly. 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 Now we turn to Fact 1. 0 892.86] /Type /Font That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. 28 0 obj 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 22 0 obj /Name /F3 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 0 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 /XHeight 430.6 869.44 511.11 597.22 830.55 894.44 575 1041.66 1169.44 894.44 319.44 0 0 0 0 0 0 /CapHeight 683.33 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 endobj Prove using the definitions that f achieves a minimum value. We needed the Extreme Value Theorem to prove Rolle’s Theorem. Theorem 1.1. 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 21 0 obj /Type /FontDescriptor /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe Proof LetA =ff(x):a •x •bg. /Subtype /Type1 /LastChar 255 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef To prove the Extreme Value Theorem, suppose a continuous function f does not achieve a maximum value on a compact set. One has to use the fact that is a real closed field, but since there are lots of real closed fields, one usually defines in a fundamentally analytic way and then proves the intermediate value theorem, which shows that is a real closed field. /StemV 80 (Extreme Value Theorem) If $f$ is a continuous function on a closed bounded interval $ [a,b],$ then $f$ must attain an absolute maximum value $f (s)$ and an absolute minimum value $f (t)$ at some numbers $s$ and $t$ in $ [a,b].$ >> << /Subtype /Type1 25 0 obj /Name /F7 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Descent -951.43 /StemV 80 /FirstChar 33 0 675.93 937.5 875 787.04 750 879.63 812.5 875 812.5 875 812.5 656.25 625 625 937.5 Proof of the Extreme Value Theorem Theorem: If f is a continuous function defined on a closed interval [a;b], then the function attains its maximum value at some point c contained in the interval. /FontDescriptor 21 0 R 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef endobj 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 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 endobj 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 /StemV 80 result for constrained problems. 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 /ItalicAngle -14 << /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Descent -250 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 /FontName /UPFELJ+CMBX10 /Type /Font /StemV 80 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 This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. Since both of these one-sided limits are equal, they must also both equal zero. 937.5 312.5 343.75 562.5 562.5 562.5 562.5 562.5 849.54 500 574.07 812.5 875 562.5 /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 >> /Descent -250 << Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 /FirstChar 33 0 0 0 0 0 0 575] /StemV 80 /FontDescriptor 9 0 R 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. It is a special case of the extremely important Extreme Value Theorem (EVT). /FontFile 17 0 R We show that, when the buyer’s values are independently distributed 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 /FontDescriptor 18 0 R 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 /Type /FontDescriptor 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 9 0 obj There are a couple of key points to note about the statement of this theorem. /StemV 80 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 $[a,b]$. /Flags 68 The Extreme Value Theorem, sometimes abbreviated EVT, says that a continuous function has a largest and smallest value on a closed interval.This is used to show thing like: There is a way to set the price of an item so as to maximize profits. 1018.52 1143.52 875 312.5 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 /CapHeight 683.33 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 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 << 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 The Mean Value Theorem for Integrals. (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. /Name /F1 /Ascent 750 892.86 892.86 892.86 1138.89 1138.89 892.86 892.86 1138.89 0 0 0 0 0 0 0 0 0 0 0 /Type /Font The Extreme Value Theorem. Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. >> /Ascent 750 /FontFile 11 0 R butions requires the proof of novel extreme value theorems for such distributions. /FontFile 26 0 R /BaseFont /UPFELJ+CMBX10 /FontName /PJRARN+CMMI10 /FirstChar 33 /FontDescriptor 24 0 R /Encoding 7 0 R Proof: There will be two parts to this proof. /FontName /JYXDXH+CMR10 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. /CapHeight 686.11 /XHeight 430.6 13 0 obj 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. >> (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 /Flags 4 /FontBBox [-116 -350 1278 850] 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 0 0 0 0 0 0 277.78] Hence by the Intermediate Value Theorem it achieves a … Indeed, complex analysis is the natural arena for such a theorem to be proven. About the Author James Lowman is an applied mathematician currently working on a Ph.D. in the field of computational fluid dynamics at the University of Waterloo. 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. /Flags 4 /Type /Encoding /Encoding 7 0 R 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 Weclaim that thereisd2[a;b]withf(d)=fi. Among all ellipses enclosing a fixed area there is one with a … /LastChar 255 /Type /FontDescriptor /ItalicAngle 0 /BaseFont /YNIUZO+CMR7 30 0 obj Suppose that is defined on the open interval and that has an absolute max at .

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