/Subtype /Type1 /StemV 80 We need Rolle’s Theorem to prove the Mean Value Theorem. 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 By the Extreme Value Theorem there must exist a value in that is a maximum. 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Sketch of Proof. /FirstChar 33 /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 /Ascent 750 /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 First we will show that there must be a ﬁnite maximum value for f (this 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 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. /StemV 80 Proof of Fermat’s Theorem. Suppose the least upper bound for \$f\$ is \$M\$. /CapHeight 683.33 /FontBBox [-134 -1122 1477 920] 24 0 obj << ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 /Subtype /Type1 endobj 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. /FontFile 23 0 R 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. /FontDescriptor 27 0 R /ItalicAngle 0 /Subtype /Type1 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. 22 0 obj 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 16 0 obj /Descent -250 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /IXTMEL+CMMI7 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 /FontName /TFBPDM+CMSY7 /Descent -250 endobj Proof of the Extreme Value Theorem. endobj /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe >> /Descent -250 We look at the proof for the upper bound and the maximum of f. /CapHeight 683.33 << The rest of the proof of this case is similar to case 2. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Also we can see that lim x → ± ∞ f (x) = ∞. f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. /StemV 80 777.78 625 916.67 750 777.78 680.56 777.78 736.11 555.56 722.22 750 750 1027.78 750 Extreme Value Theorem: If a function 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. 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 %PDF-1.3 Suppose there is no such \$c\$. 3 /LastChar 255 For every ε > 0. /CapHeight 683.33 /Type /Font /CapHeight 686.11 519.84 668.05 592.71 661.99 526.84 632.94 686.91 713.79 755.96 0 0 0 0 0 0 0 0 0 First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] /Encoding 7 0 R 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. 21 0 obj /FontFile 26 0 R /ItalicAngle -14 If a function \$f\$ is continuous on \$[a,b]\$, then it attains its maximum and minimum values on \$[a,b]\$. 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 It is necessary to find a point d in [ a , b ] such that M = f ( d ). /Flags 4 0 0 0 0 0 0 277.78] 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 k – ε < f (c) < k + ε. 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 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. 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi 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. This is the Weierstrass Extreme Value Theorem. 25 0 obj Now we turn to Fact 1. We needed the Extreme Value Theorem to prove Rolle’s Theorem. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << 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. /Subtype /Type1 /ItalicAngle -14 0 892.86] result for constrained problems. 18 0 obj >> The Mean Value Theorem for Integrals. 15 0 obj 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. /Type /Font /Type /FontDescriptor The Extreme Value Theorem. >> 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]\$. /Type /Font We show that, when the buyer’s values are independently distributed Weclaim that thereisd2[a;b]withf(d)=ﬁ. 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 0 693.75 954.37 868.93 797.62 844.5 935.62 886.31 677.58 769.84 716.89 880.04 742.68 << /FontBBox [-119 -350 1308 850] >> /Type /FontDescriptor 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 As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. 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