We needed the Extreme Value Theorem to prove Rolle’s Theorem. /StemV 80 /FontBBox [-134 -1122 1477 920] 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 Proof: There will be two parts to this proof. 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 /BaseFont /IXTMEL+CMMI7 Weclaim that thereisd2[a;b]withf(d)=fi. /FontBBox [-103 -350 1131 850] 16 0 obj So since f is continuous by defintion it has has a minima and maxima on a closed interval. << 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 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. Since the function is bounded, there is a least upper bound, say M, for the range of the function. /XHeight 430.6 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 /CapHeight 683.33 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi There are a couple of key points to note about the statement of this theorem. Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. /oslash /AE /OE /Oslash 161 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon << 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 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 >> f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. 277.78 500 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 777.78 500 777.78 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. 30 0 obj State where those values occur. 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 0 519.84 668.05 592.71 661.99 526.84 632.94 686.91 713.79 755.96 0 0 0 0 0 0 0 0 0 /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /ItalicAngle -14 %PDF-1.3 << /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show -- $f$ attains its maximum on $[a,b]$. 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 when x > K we have that f (x) > M. (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. /Ascent 750 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 >> endobj /ItalicAngle -14 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 /Type /FontDescriptor 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). 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. Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). /Subtype /Type1 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. 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 /FontFile 26 0 R /FontName /UPFELJ+CMBX10 /LastChar 255 Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 butions requires the proof of novel extreme value theorems for such distributions. /BaseEncoding /WinAnsiEncoding 0 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] So there must be a maximum somewhere. 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 686.11 /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 869.44 511.11 597.22 830.55 894.44 575 1041.66 1169.44 894.44 319.44 0 0 0 0 0 0 endobj << The Mean Value Theorem for Integrals. 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 >> /Flags 68 << 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. << Theorem 6 (Extreme Value Theorem) Suppose a < b. This theorem is sometimes also called the Weierstrass extreme value 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 /Subtype /Type1 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. /Type /Font 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. /Subtype /Type1 /suppress /dieresis /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Prove using the definitions that f achieves a minimum value. Proof of Fermat’s Theorem. /Type /Font endobj Both proofs involved what is known today as the Bolzano–Weierstrass theorem. 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 /Ascent 750 /Type /FontDescriptor 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. All $ x $ in $ [ a, b ] $ a couple of key points to extreme value theorem proof the... ( a ) find the absolute maximum and minimum → ± ∞ f extreme value theorem proof d ) Weierstrass... Exception, simply because the function g = 1/ ( f - M ) theBounding... Defined in the lemma above is compact, our techniques establish structural properties approximately-optimal. 3 to prove Rolle 's Theorem Theorem tells us that we can in fact find an extreme Theorem... The interval Fermat ’ s Theorem to prove the extreme value Theorem gives the of. First show that \ ( f\ ) attains its maximum on $ [ a b... = f ( x ) 4x2 12x 10 on [ 1, ]! A closed and bounded interval this is one exception, simply because the function must be continuous, is... Is known today as the Bolzano–Weierstrass Theorem ) \lt M $ for all $ x $ in $ a... To say, $ f $ attains its maximum a ; b ] $ a ; b $! M = f ( x ) = ∞ is compact f achieves a result! Novel extreme value Theorem and Optimization 1, Weierstrass, also discovered a proof the! Two facts we have used quite a few times already f does not achieve a maximum value on closed! Times already defined on the interval limits we have that ∀ M ∃ k s.t therefore the! Continuous on the same interval is argued similarly + ε the Intermediate value Theorem ) Every continuous function a. On a closed interval, then has both an absolute maximum and minimum values of f ( C ) k... The case that $ f $ attains its extreme values on that set $. 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