Both proofs involved what is known today as the Bolzano–Weierstrass theorem. 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 0 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] Typically, it is proved in a course on real analysis. 0 0 0 0 0 0 277.78] 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 Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 Since the function is bounded, there is a least upper bound, say M, for the range of the function. Now we turn to Fact 1. endobj /Type /Font /ItalicAngle 0 16 0 obj Sketch of Proof. /Type /FontDescriptor /Flags 4 Among all ellipses enclosing a fixed area there is one with a … /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 /Type /Font Hence, the theorem is proved. Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 stream << https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem endobj Theorem 1.1. It is necessary to find a point d in [ a , b ] such that M = f ( d ). 18 0 obj endobj >> Thus for all in . >> >> Examples 7.4 – The Extreme Value Theorem and Optimization 1. /FontDescriptor 12 0 R /FontName /IXTMEL+CMMI7 12 0 obj 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 0 575] 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 $[a,b]$. 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 /FontDescriptor 18 0 R << Suppose the least upper bound for $f$ is $M$. /FontBBox [-114 -350 1253 850] We show that, when the buyer’s values are independently distributed /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] /Type /Font Therefore proving Fermat’s Theorem for Stationary Points. /oslash /AE /OE /Oslash 161 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon First we will show that there must be a ﬁnite maximum value for f (this /suppress /dieresis /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Flags 68 << /Flags 4 /Encoding 7 0 R /Descent -250 /FirstChar 33 /Ascent 750 869.44 511.11 597.22 830.55 894.44 575 1041.66 1169.44 894.44 319.44 0 0 0 0 0 0 Since f never attains the value M, g is continuous, and is therefore itself bounded. /FontBBox [-100 -350 1100 850] /Type /FontDescriptor 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. 15 0 obj 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent /Descent -250 >> /Name /F6 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 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /FontName /PJRARN+CMMI10 Theorem 6 (Extreme Value Theorem) Suppose a < b. 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. /LastChar 255 /FontName /UPFELJ+CMBX10 777.78 625 916.67 750 777.78 680.56 777.78 736.11 555.56 722.22 750 750 1027.78 750 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 We needed the Extreme Value Theorem to prove Rolle’s Theorem. To prove the Extreme Value Theorem, suppose a continuous function f does not achieve a maximum value on a compact set. The Extreme Value Theorem. /CapHeight 686.11 Therefore by the definition of limits we have that ∀ M ∃ K s.t. 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 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. /CapHeight 683.33 569.45] /BaseFont /YNIUZO+CMR7 The Mean Value Theorem for Integrals. Hence by the Intermediate Value Theorem it achieves a … /FirstChar 33 /Type /FontDescriptor 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. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Flags 4 30 0 obj 22 0 obj result for constrained problems. /ItalicAngle -14 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 Proof of Fermat’s Theorem. << /FontDescriptor 27 0 R The proof of the extreme value theorem is beyond the scope of this text. 277.78 500 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 777.78 500 777.78 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 So there must be a maximum somewhere. /Flags 4 A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). /Encoding 7 0 R /Flags 68 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta Proof: There will be two parts to this proof. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500] Suppose that is defined on the open interval and that has an absolute max at . (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. /LastChar 255 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]. 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 The result was also discovered later by Weierstrass in 1860. /FontFile 11 0 R Since both of these one-sided limits are equal, they must also both equal zero. /FontBBox [-116 -350 1278 850] 9 0 obj As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. /Type /FontDescriptor 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. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 19 0 obj 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. endobj /StemV 80 819.39 934.07 838.69 724.51 889.43 935.62 506.3 632.04 959.93 783.74 1089.39 904.87 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. This makes sense because the function must go up (as) and come back down to where it started (as). 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. /FontName /YNIUZO+CMR7 /XHeight 430.6 endobj endobj Indeed, complex analysis is the natural arena for such a theorem to be proven. Suppose there is no such $c$. 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. /Ascent 750 /Descent -250 /Subtype /Type1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 493.98 437.5 570.03 517.02 571.41 437.15 540.28 595.83 625.69 651.39 0 0 0 0 0 0 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /BaseFont /UPFELJ+CMBX10 (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. /FontBBox [-115 -350 1266 850] /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. /FontName /TFBPDM+CMSY7 /FontName /NRFPYP+CMBX12 /XHeight 444.4 /Flags 68 28 0 obj /Name /F2 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. /Subtype /Type1 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 << We now build a basic existence result for unconstrained problems based on this theorem. 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 Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. /StemV 80 937.5 312.5 343.75 562.5 562.5 562.5 562.5 562.5 849.54 500 574.07 812.5 875 562.5 << 767.86 876.99 829.37 630.96 815.48 843.26 843.26 1150.8 843.26 843.26 692.46 323.41 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Proof LetA =ff(x):a •x •bg. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /Ascent 750 /Ascent 750 575 638.89 606.94 473.61 453.61 447.22 638.89 606.94 830.55 606.94 606.94 511.11 << /StemV 80 /LastChar 255 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 27 0 obj %PDF-1.3 24 0 obj /FontFile 23 0 R endobj << 7 0 obj /FirstChar 33 (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 << /Subtype /Type1 << endobj /StemV 80 /Type /Font 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. butions requires the proof of novel extreme value theorems for such distributions. << 647.77 600.08 519.25 476.14 519.84 588.6 544.15 422.79 668.82 677.58 694.62 572.76 Prove using the definitions that f achieves a minimum value. State where those values occur. /LastChar 255 >> If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show -- $f$ attains its maximum on $[a,b]$. 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. 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /FontBBox [-103 -350 1131 850] 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 /ItalicAngle 0 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 That is to say, $f$ attains its maximum on $[a,b]$. Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. We look at the proof for the upper bound and the maximum of f. endobj 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /StemV 80 We will ﬁrst show that \(f\) attains its maximum. /StemV 80 864.58 849.54 1162.04 849.54 849.54 687.5 312.5 581.02 312.5 562.5 312.5 312.5 546.88 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi 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. 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 /Name /F1 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 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 ]. 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. 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 when x > K we have that f (x) > M. /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 The extreme value theorem is used to prove Rolle's theorem. endobj We need Rolle’s Theorem to prove the Mean Value Theorem. We can choose the value to be that maximum. By the Extreme Value Theorem there must exist a value in that is a maximum. /CapHeight 683.33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. /Ascent 750 /BaseFont /NRFPYP+CMBX12 So since f is continuous by defintion it has has a minima and maxima on a closed interval. /FontFile 17 0 R 21 0 obj 0 892.86] 0 0 0 339.29] endobj 462.3 462.3 462.3 1138.89 1138.89 478.18 619.66 502.38 510.54 594.7 542.02 557.05 769.85 769.85 892.86 892.86 523.81 523.81 523.81 708.34 892.86 892.86 892.86 0 892.86 /Subtype /Type1 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). endobj 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 /FontFile 26 0 R 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. /Descent -250 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 /XHeight 444.4 /Widths [342.59 581.02 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. /Descent -951.43 For every ε > 0. The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. /ItalicAngle 0 25 0 obj /Descent -250 /FontFile 14 0 R 10 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 k – ε < f (c) < k + ε. /LastChar 255 /Filter [/FlateDecode] /Ascent 750 /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 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. There are a couple of key points to note about the statement of this theorem. State where those values occur. >> 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. /StemV 80 /CapHeight 683.33 >> (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].$ << /BaseFont /PJRARN+CMMI10 /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 /Encoding 7 0 R /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 892.86 892.86 892.86 1138.89 1138.89 892.86 892.86 1138.89 0 0 0 0 0 0 0 0 0 0 0 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 the Weierstrass Extreme Value Theorem. 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 /FontBBox [-134 -1122 1477 920] We prove the case that $f$ attains its maximum value on $[a,b]$. /BaseEncoding /WinAnsiEncoding Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 3 /Subtype /Type1 /Type /Font endobj /XHeight 430.6 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. Weclaim that thereisd2[a;b]withf(d)=ﬁ. >> 511.11 638.89 527.08 351.39 575 638.89 319.44 351.39 606.94 319.44 958.33 638.89 >> >> /Type /Font 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /FontFile 8 0 R 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 /Name /F5 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 /BaseFont /IXTMEL+CMMI7 /FirstChar 33 /Subtype /Type1 /Type /FontDescriptor /FontDescriptor 9 0 R 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /Type /Encoding /FontFile 20 0 R 13 0 obj If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. /FontBBox [-119 -350 1308 850] >> 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 The proof that $f$ attains its minimum on the same interval is argued similarly. Theorem 7.3 (Mean Value Theorem MVT). /ItalicAngle -14 /Name /F7 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 /Widths [323.41 569.45 938.5 569.45 938.5 876.99 323.41 446.43 446.43 569.45 876.99 /ItalicAngle -14 f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. /LastChar 255 >> /Type /FontDescriptor /FontDescriptor 15 0 R /ItalicAngle 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). /Name /F3 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! 312.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 /BaseFont /JYXDXH+CMR10 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 >> /LastChar 255 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 /CapHeight 686.11 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 /Name /F4 Then the image D as defined in the lemma above is compact. /FirstChar 33 It is a special case of the extremely important Extreme Value Theorem (EVT). << /BaseFont /TFBPDM+CMSY7 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]. 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 Letﬁ =supA. /CapHeight 683.33 Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. 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. The rest of the proof of this case is similar to case 2. 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 /CapHeight 683.33 1138.89 339.29 339.29 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /Type /FontDescriptor 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. endobj /Encoding 7 0 R The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. 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 Sketch of Proof. 625 500 625 513.31 343.75 562.5 625 312.5 343.75 593.75 312.5 937.5 625 562.5 625 /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring Also we can see that lim x → ± ∞ f (x) = ∞. /FontDescriptor 24 0 R /XHeight 430.6 << Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. /FontDescriptor 21 0 R /Length 3528 This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. 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