(Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. endobj Among all ellipses enclosing a fixed area there is one with a … (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. /CapHeight 683.33 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 stream 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. /Type /FontDescriptor 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 /LastChar 255 /FontName /NRFPYP+CMBX12 /StemV 80 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. 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 << 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]. /Name /F4 There are a couple of key points to note about the statement of this theorem. 27 0 obj 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 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. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /FontDescriptor 15 0 R By the Extreme Value Theorem there must exist a value in that is a maximum. /FontBBox [-114 -350 1253 850] /LastChar 255 For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 /FontFile 26 0 R >> Then the image D as defined in the lemma above is compact. /BaseFont /YNIUZO+CMR7 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 /Type /FontDescriptor >> /FontName /YNIUZO+CMR7 0 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] Hence by the Intermediate Value Theorem it achieves a … Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. The rest of the proof of this case is similar to case 2. Thus for all in . << Letfi =supA. We show that, when the buyer’s values are independently distributed << The proof that $f$ attains its minimum on the same interval is argued similarly. 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 Therefore proving Fermat’s Theorem for Stationary Points. /BaseFont /TFBPDM+CMSY7 ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. /Descent -250 /Type /Font /Type /Font /FontFile 11 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 Weclaim that thereisd2[a;b]withf(d)=fi. 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 /FontDescriptor 18 0 R << /FirstChar 33 /XHeight 444.4 937.5 312.5 343.75 562.5 562.5 562.5 562.5 562.5 849.54 500 574.07 812.5 875 562.5 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 /Ascent 750 /Flags 68 21 0 obj /BaseFont /UPFELJ+CMBX10 /Flags 68 /Encoding 7 0 R << 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. This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. /FontDescriptor 9 0 R /Type /FontDescriptor >> 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 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! /FontName /JYXDXH+CMR10 /Name /F1 endobj 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 %PDF-1.3 472.22 472.22 777.78 750 708.34 722.22 763.89 680.56 652.78 784.72 750 361.11 513.89 /ItalicAngle 0 For every ε > 0. 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. State where those values occur. 0 0 0 339.29] 819.39 934.07 838.69 724.51 889.43 935.62 506.3 632.04 959.93 783.74 1089.39 904.87 Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. /FontBBox [-119 -350 1308 850] /FontDescriptor 12 0 R Proof of the Extreme Value Theorem. 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 /FontBBox [-103 -350 1131 850] 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 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 $[a,b]$. 15 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 >> /FontBBox [-115 -350 1266 850] 25 0 obj In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. 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]. 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). Both proofs involved what is known today as the Bolzano–Weierstrass theorem. /Descent -250 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 19 0 obj /Length 3528 The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. 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 Fermat’s Theorem. /Flags 4 /Filter [/FlateDecode] Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). We need Rolle’s Theorem to prove the Mean Value Theorem. 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). /FontName /PJRARN+CMMI10 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. 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 /BaseFont /JYXDXH+CMR10 Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. 0 0 0 0 0 0 575] >> 569.45] 10 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 The result was also discovered later by Weierstrass in 1860. /LastChar 255 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. Typically, it is proved in a course on real analysis. Then $f(x) \lt M$ for all $x$ in $[a,b]$. /CapHeight 683.33 endobj 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500] Proof: There will be two parts to this proof. 30 0 obj /CapHeight 686.11 (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. /Subtype /Type1 /LastChar 255 Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. 0 892.86] 7 0 obj 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 /Ascent 750 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. The Extreme Value Theorem. /Ascent 750 endobj /Subtype /Type1 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. >> 22 0 obj /XHeight 430.6 /BaseFont /IXTMEL+CMMI7 Consider the function g = 1/ (f - M). 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 will first show that \(f\) attains its maximum. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Type /FontDescriptor >> We prove the case that $f$ attains its maximum value on $[a,b]$. (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].$ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 << /Flags 4 /Descent -250 /FontFile 17 0 R 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 This is the Weierstrass Extreme Value Theorem. Now we turn to Fact 1. So since f is continuous by defintion it has has a minima and maxima on a closed interval. 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. 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 /ItalicAngle -14 Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). Theorem 7.3 (Mean Value Theorem MVT). endobj 12 0 obj /Descent -250 Proof LetA =ff(x):a •x •bg. endobj /CapHeight 683.33 /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 /FontDescriptor 21 0 R 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. /StemV 80 << /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring 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 endobj 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 /Encoding 7 0 R /FontFile 8 0 R 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 /StemV 80 /Type /Font butions requires the proof of novel extreme value theorems for such distributions. /XHeight 430.6 /StemV 80 /FirstChar 33 >> /StemV 80 575 638.89 606.94 473.61 453.61 447.22 638.89 606.94 830.55 606.94 606.94 511.11 /Encoding 7 0 R /Type /Font Examples 7.4 – The Extreme Value Theorem and Optimization 1. 511.11 638.89 527.08 351.39 575 638.89 319.44 351.39 606.94 319.44 958.33 638.89 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 /LastChar 255 /Ascent 750 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Type /Font /Ascent 750 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 -14 k – ε < f (c) < k + ε. endobj /ItalicAngle 0 28 0 obj 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. /Subtype /Type1 First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). >> 462.3 462.3 462.3 1138.89 1138.89 478.18 619.66 502.38 510.54 594.7 542.02 557.05 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 We can choose the value to be that maximum. To prove the Extreme Value Theorem, suppose a continuous function f does not achieve a maximum value on a compact set. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] endobj State where those values occur. Also we can see that lim x → ± ∞ f (x) = ∞. 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 << The proof of the extreme value theorem is beyond the scope of this text. >> 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /BaseEncoding /WinAnsiEncoding /Name /F2 /Descent -951.43 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 We look at the proof for the upper bound and the maximum of f. 647.77 600.08 519.25 476.14 519.84 588.6 544.15 422.79 668.82 677.58 694.62 572.76 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 Since both of these one-sided limits are equal, they must also both equal zero. /BaseFont /PJRARN+CMMI10 /Flags 4 /suppress /dieresis /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 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. << 0 0 0 0 0 0 277.78] /FontBBox [-116 -350 1278 850] /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 << Prove using the definitions that f achieves a minimum value. /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 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. /LastChar 255 endobj 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. Theorem 6 (Extreme Value Theorem) Suppose a < b. 16 0 obj 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 The Mean Value Theorem for Integrals. /Subtype /Type1 First we will show that there must be a finite maximum value for f (this << Therefore by the definition of limits we have that ∀ M ∃ K s.t. >> f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. endobj /Encoding 7 0 R The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. 13 0 obj 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 /Subtype /Type1 Suppose the least upper bound for $f$ is $M$. /StemV 80 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 1138.89 339.29 339.29 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 That is to say, $f$ attains its maximum on $[a,b]$. 277.78 500 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 777.78 500 777.78 The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 /FontFile 20 0 R /XHeight 430.6 /Widths [323.41 569.45 938.5 569.45 938.5 876.99 323.41 446.43 446.43 569.45 876.99 Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. when x > K we have that f (x) > M. /XHeight 430.6 /FontBBox [-134 -1122 1477 920] /ItalicAngle 0 << /FontDescriptor 24 0 R /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 It is a special case of the extremely important Extreme Value Theorem (EVT). /FontBBox [-100 -350 1100 850] /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 /ItalicAngle -14 593.75 459.49 443.75 437.5 625 593.75 812.5 593.75 593.75 500 562.5 1125 562.5 562.5 /Descent -250 /Name /F7 /Ascent 750 /FirstChar 33 24 0 obj /FontName /TFBPDM+CMSY7 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 0 646.83 646.83 769.85 585.32 831.35 831.35 892.86 892.86 708.34 917.6 753.44 620.18 >> >> /FontName /UPFELJ+CMBX10 Sketch of Proof. If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show -- $f$ attains its maximum on $[a,b]$. 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 /FontFile 14 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi << /Widths [342.59 581.02 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 /LastChar 255 /CapHeight 683.33 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 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. If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. endobj The extreme value theorem is used to prove Rolle's theorem. /Type /Encoding /CapHeight 683.33 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 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. result for constrained problems. /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 This theorem is sometimes also called the Weierstrass extreme value 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. /Flags 68 As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright >> /Descent -250 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. 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. /Name /F6 >> 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 If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. /CapHeight 686.11 >> /FirstChar 33 /XHeight 444.4 /oslash /AE /OE /Oslash 161 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Name /F3 Indeed, complex analysis is the natural arena for such a theorem to be proven. endobj If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /Type /FontDescriptor 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 /FirstChar 33 /FirstChar 33 Since the function is bounded, there is a least upper bound, say M, for the range of the function. /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /BaseFont /NRFPYP+CMBX12 769.85 769.85 769.85 769.85 708.34 708.34 523.81 523.81 523.81 523.81 585.32 585.32 /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 << https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem 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 Suppose that is defined on the open interval and that has an absolute max at . This makes sense because the function must go up (as) and come back down to where it started (as). It is necessary to find a point d in [ a , b ] such that M = f ( d ). We now build a basic existence result for unconstrained problems based on this theorem. /Ascent 750 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 /Name /F5 /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe 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 ]. /FontDescriptor 27 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 /FontFile 23 0 R /ItalicAngle 0 /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta 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 Suppose there is no such $c$. /Subtype /Type1 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /Type /FontDescriptor /Type /FontDescriptor 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent 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. /Type /Font Theorem 1.1. /StemV 80 So there must be a maximum somewhere. 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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. /XHeight 430.6 Hence, the theorem is proved. /Subtype /Type1 Sketch of Proof. 18 0 obj << /Flags 4 M = f ( x ) 4x2 12x 10 on [ 1 3! 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