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. /ItalicAngle 0 Theorem 6 (Extreme Value Theorem) Suppose a < b. There are a couple of key points to note about the statement of this theorem. Hence, the theorem is proved. The Extreme Value Theorem. 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 ]. /CapHeight 683.33 >> endobj The extreme value theorem itself was first proved by the Bohemian mathematician and philosopher Bernard Bolzano in 1830, but his book, Function Theory, was only published a hundred years later in 1930. The rest of the proof of this case is similar to case 2. 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 Suppose there is no such $c$. /FirstChar 33 /Descent -951.43 That is to say, $f$ attains its maximum on $[a,b]$. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 562.5] Proof of the Extreme Value Theorem. Since both of these one-sided limits are equal, they must also both equal zero. /Ascent 750 If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show -- $f$ attains its maximum on $[a,b]$. << << /Descent -250 It is a special case of the extremely important Extreme Value Theorem (EVT). 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). /StemV 80 /Ascent 750 /StemV 80 << << /BaseEncoding /WinAnsiEncoding /BaseFont /JYXDXH+CMR10 /FontName /PJRARN+CMMI10 /Ascent 750 /oslash /AE /OE /Oslash 161 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon 16 0 obj << This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. 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 /FontName /YNIUZO+CMR7 That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. /FirstChar 33 892.86 892.86 892.86 1138.89 1138.89 892.86 892.86 1138.89 0 0 0 0 0 0 0 0 0 0 0 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Among all ellipses enclosing a fixed area there is one with a … /Descent -250 /StemV 80 /BaseFont /PJRARN+CMMI10 endobj /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /BaseFont /IXTMEL+CMMI7 endobj Then the image D as defined in the lemma above is compact. /Encoding 7 0 R /FontFile 8 0 R It is necessary to find a point d in [ a , b ] such that M = f ( d ). /FontBBox [-134 -1122 1477 920] Thus for all in . 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 Consider the function g = 1/ (f - M). 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 /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 /BaseFont /UPFELJ+CMBX10 625 500 625 513.31 343.75 562.5 625 312.5 343.75 593.75 312.5 937.5 625 562.5 625 Suppose the least upper bound for $f$ is $M$. This is the Weierstrass Extreme Value Theorem. stream 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 312.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 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 24 0 obj /Type /Font Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. Also we can see that lim x → ± ∞ f (x) = ∞. endobj /LastChar 255 >> /LastChar 255 If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. /Subtype /Type1 /FontDescriptor 12 0 R endobj /Subtype /Type1 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 Now we turn to Fact 1. /Flags 4 We needed the Extreme Value Theorem to prove Rolle’s Theorem. /Name /F6 /FontFile 17 0 R /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi 0 0 0 0 0 0 575] 19 0 obj We prove the case that $f$ attains its maximum value on $[a,b]$. /Subtype /Type1 647.77 600.08 519.25 476.14 519.84 588.6 544.15 422.79 668.82 677.58 694.62 572.76 27 0 obj endobj 22 0 obj /ItalicAngle 0 Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. /FontBBox [-116 -350 1278 850] Therefore proving Fermat’s Theorem for Stationary Points. /XHeight 430.6 769.85 769.85 892.86 892.86 523.81 523.81 523.81 708.34 892.86 892.86 892.86 0 892.86 The result was also discovered later by Weierstrass in 1860. /BaseFont /TFBPDM+CMSY7 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Proof LetA =ff(x):a •x •bg. 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 endobj First we will show that there must be a ﬁnite maximum value for f (this 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]. /Widths [342.59 581.02 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 1013.89 777.78 277.78 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 >> /FirstChar 33 /Type /FontDescriptor /FontName /UPFELJ+CMBX10 /Type /Font 462.3 462.3 462.3 1138.89 1138.89 478.18 619.66 502.38 510.54 594.7 542.02 557.05 >> /Type /Font /Type /Encoding The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 /FontFile 14 0 R /Type /FontDescriptor Examples 7.4 – The Extreme Value Theorem and Optimization 1. 819.39 934.07 838.69 724.51 889.43 935.62 506.3 632.04 959.93 783.74 1089.39 904.87 A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). 769.85 769.85 769.85 769.85 708.34 708.34 523.81 523.81 523.81 523.81 585.32 585.32 /Subtype /Type1 /BaseFont /NRFPYP+CMBX12 The Mean Value Theorem for Integrals. State where those values occur. By the Extreme Value Theorem there must exist a value in that is a maximum. Since the function is bounded, there is a least upper bound, say M, for the range of the function. /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 Theorem 7.3 (Mean Value Theorem MVT). 446.43 630.96 600.2 815.48 600.2 600.2 507.94 569.45 1138.89 569.45 569.45 0 706.35 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 endobj Indeed, complex analysis is the natural arena for such a theorem to be proven. 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 0 892.86] The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. << The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. /FontName /JYXDXH+CMR10 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef endobj 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. 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 << 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 The extreme value theorem is used to prove Rolle's theorem. Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. The proof that $f$ attains its minimum on the same interval is argued similarly. << 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /Type /FontDescriptor 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]. 569.45 815.48 876.99 569.45 1013.89 1136.91 876.99 323.41 0 0 0 0 0 0 0 0 0 0 0 0 The proof of the extreme value theorem is beyond the scope of this text. As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. /LastChar 255 /FontFile 26 0 R /FontFile 11 0 R (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].$ /LastChar 255 /Name /F3 /Descent -250 Therefore by the definition of limits we have that ∀ M ∃ K s.t. /Ascent 750 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 /FontDescriptor 27 0 R << This theorem is sometimes also called the Weierstrass extreme value theorem. /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 >> 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). /Length 3528 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 Sketch of Proof. >> endobj endobj endobj 0 0 0 339.29] /Name /F1 We now build a basic existence result for unconstrained problems based on this theorem. 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. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] << 12 0 obj /XHeight 430.6 In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 >> << 15 0 obj Prove using the definitions that f achieves a minimum value. /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega We show that, when the buyer’s values are independently distributed /LastChar 255 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 767.86 876.99 829.37 630.96 815.48 843.26 843.26 1150.8 843.26 843.26 692.46 323.41 /FontName /IXTMEL+CMMI7 /StemV 80 /FontBBox [-115 -350 1266 850] (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 569.45] State where those values occur. Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. << Typically, it is proved in a course on real analysis. /Descent -250 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 13 0 obj /Flags 4 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 /Flags 68 /Encoding 7 0 R 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. 511.11 638.89 527.08 351.39 575 638.89 319.44 351.39 606.94 319.44 958.33 638.89 /FontName /TFBPDM+CMSY7 /FontDescriptor 9 0 R /CapHeight 683.33 Hence by the Intermediate Value Theorem it achieves a … /FontName /NRFPYP+CMBX12 /FontDescriptor 24 0 R /Ascent 750 result for constrained problems. /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. For every ε > 0. 277.78 500 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 777.78 500 777.78 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. First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). << 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. 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. /Flags 4 /FontFile 20 0 R endobj 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 /FirstChar 33 /Subtype /Type1 /Encoding 7 0 R 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. 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. Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. << We need Rolle’s Theorem to prove the Mean Value Theorem. k – ε < f (c) < k + ε. 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 $[a,b]$. /StemV 80 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 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. /ItalicAngle -14 This makes sense because the function must go up (as) and come back down to where it started (as). Weclaim that thereisd2[a;b]withf(d)=ﬁ. endobj 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 /Name /F2 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 21 0 obj For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). /FontBBox [-119 -350 1308 850] 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 /Font /LastChar 255 (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. 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. Unformatted text preview: Intermediate Value Property for Functions with Antiderivatives Theorem 5.3. let Suppose that ϕ is di erentiable at each point of the interval I and ϕ (x) = f (x) for all x ∈ I. /Flags 68 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 /Subtype /Type1 /Type /FontDescriptor >> /Ascent 750 Extreme Value Theorem: If a function is continuous in a closed interval , with the maximum of at and the minimum of at then and are critical values of Proof: The proof follows from Fermat’s theorem and is left as an exercise for the student. 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 >> /ItalicAngle 0 /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 Then $f(x) \lt M$ for all $x$ in $[a,b]$. /XHeight 444.4 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 /BaseFont /YNIUZO+CMR7 /Encoding 7 0 R << %PDF-1.3 /FontFile 23 0 R 864.58 849.54 1162.04 849.54 849.54 687.5 312.5 581.02 312.5 562.5 312.5 312.5 546.88 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 0 0 646.83 646.83 769.85 585.32 831.35 831.35 892.86 892.86 708.34 917.6 753.44 620.18 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. Since f never attains the value M, g is continuous, and is therefore itself bounded. 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 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 /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. >> /Type /Font /FirstChar 33 /Filter [/FlateDecode] https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. 0 0 0 0 0 0 277.78] /XHeight 444.4 /CapHeight 683.33 /FontDescriptor 15 0 R /XHeight 430.6 endobj ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. 575 638.89 606.94 473.61 453.61 447.22 638.89 606.94 830.55 606.94 606.94 511.11 >> /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring /Widths [323.41 569.45 938.5 569.45 938.5 876.99 323.41 446.43 446.43 569.45 876.99 /Subtype /Type1 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 3 /CapHeight 686.11 /CapHeight 686.11 /FirstChar 33 772.4 639.7 565.63 517.73 444.44 405.9 437.5 496.53 469.44 353.94 576.16 583.34 602.55 Theorem 1.1. 18 0 obj 472.22 472.22 777.78 750 708.34 722.22 763.89 680.56 652.78 784.72 750 361.11 513.89 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. >> 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 /FontDescriptor 18 0 R 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! 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500] /ItalicAngle -14 /StemV 80 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /StemV 80 /FontBBox [-100 -350 1100 850] We look at the proof for the upper bound and the maximum of f. /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 /suppress /dieresis /.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. Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 /Name /F4 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 /Type /FontDescriptor /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Type /FontDescriptor We can choose the value to be that maximum. Proof of Fermat’s Theorem. /Flags 68 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 /Type /Font 1138.89 339.29 339.29 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 /CapHeight 683.33 butions requires the proof of novel extreme value theorems for such distributions. 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 /Ascent 750 >> So since f is continuous by defintion it has has a minima and maxima on a closed interval. /Flags 4 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 /CapHeight 683.33 /Descent -250 /LastChar 255 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. 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 /FontBBox [-103 -350 1131 850] So there must be a maximum somewhere. 7 0 obj /FirstChar 33 f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. /Name /F7 /Descent -250 9 0 obj >> Sketch of Proof. /FontDescriptor 21 0 R 777.78 625 916.67 750 777.78 680.56 777.78 736.11 555.56 722.22 750 750 1027.78 750 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. 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. Suppose that is defined on the open interval and that has an absolute max at . 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 /XHeight 430.6 /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright We will ﬁrst show that \(f\) attains its maximum. 937.5 312.5 343.75 562.5 562.5 562.5 562.5 562.5 849.54 500 574.07 812.5 875 562.5 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 /Type /Font << 30 0 obj Proof: There will be two parts to this proof. >> /Name /F5 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 28 0 obj /FontBBox [-114 -350 1253 850] 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 25 0 obj 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 Both proofs involved what is known today as the Bolzano–Weierstrass theorem. /ItalicAngle 0 when x > K we have that f (x) > M. 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