
On January 19, 2021
extreme value theorem proof
/LastChar 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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. 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 k – ε < f (c) < k + ε. Also we can see that lim x → ± ∞ f (x) = ∞. Theorem 7.3 (Mean Value Theorem MVT). 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 [100 350 1100 850] /Encoding 7 0 R 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 Hence, the theorem is proved. /ItalicAngle 0 /Name /F4 >> 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 Weclaim that thereisd2[a;b]withf(d)=ﬁ. /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 << 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 /FontDescriptor 21 0 R 13 0 obj Suppose there is no such $c$. /Encoding 7 0 R As a byproduct, our techniques establish structural properties of approximatelyoptimal and nearoptimal solutions. /Filter [/FlateDecode] butions requires the proof of novel extreme value theorems for such distributions. If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. 0 892.86] /FirstChar 33 /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring stream 0 0 0 0 0 0 277.78] 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 0 0 0 0 0 0 0 0 0 0 0 Indeed, complex analysis is the natural arena for such a theorem to be proven. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 %PDF1.3 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 /FontDescriptor 12 0 R /ItalicAngle 14 >> /FontFile 26 0 R 16 0 obj /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef Since the function is bounded, there is a least upper bound, say M, for the range of the function. Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{Mf(x)}}$. /Flags 4 Suppose that is defined on the open interval and that has an absolute max at . Then the image D as defined in the lemma above is compact. The Extreme Value Theorem. /Descent 250 /Descent 250 Consider the function g = 1/ (f  M). /Name /F3 /FontDescriptor 18 0 R Therefore proving Fermat’s Theorem for Stationary Points. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /suppress /dieresis /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 /Encoding 7 0 R 0 0 0 0 0 0 575] /Type /Font First we will show that there must be a ﬁnite maximum value for f (this So there must be a maximum somewhere. endobj endobj /Type /Font 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 /CapHeight 683.33 State where those values occur. Theorem 6 (Extreme Value Theorem) Suppose a < b. which implies (upon multiplication of both sides by the positive $Mf(x)$, 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! 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 0 562.5] /.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 /dotlessj /grave /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 >> Typically, it is proved in a course on real analysis. 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 $[a,b]$. 767.86 876.99 829.37 630.96 815.48 843.26 843.26 1150.8 843.26 843.26 692.46 323.41 /XHeight 430.6 Since f never attains the value M, g is continuous, and is therefore itself bounded. /Type /FontDescriptor f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. 15 0 obj /FontName /JYXDXH+CMR10 A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). /Flags 4 There are a couple of key points to note about the statement of this theorem. /StemV 80 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 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 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. /Name /F5 ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. /FontDescriptor 27 0 R 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. This makes sense because the function must go up (as) and come back down to where it started (as). (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. endobj /CapHeight 683.33 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. /FontName /IXTMEL+CMMI7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 The rest of the proof of this case is similar to case 2. 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 /StemV 80 It is necessary to find a point d in [ a , b ] such that M = f ( d ). 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 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. /LastChar 255 >> 21 0 obj /Length 3528 /Subtype /Type1 /BaseFont /IXTMEL+CMMI7 The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. 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 /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. /LastChar 255 >> 7 0 obj 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 593.75 459.49 443.75 437.5 625 593.75 812.5 593.75 593.75 500 562.5 1125 562.5 562.5 Among all ellipses enclosing a fixed area there is one with a … For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright << We can choose the value to be that maximum. /FontName /TFBPDM+CMSY7 The extreme value theorem is used to prove Rolle's 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 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]. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef We need Rolle’s Theorem to prove the Mean Value Theorem. /Widths [342.59 581.02 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded 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 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. 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 endobj Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). /BaseEncoding /WinAnsiEncoding << /FontBBox [134 1122 1477 920] >> /StemV 80 << The proof that $f$ attains its minimum on the same interval is argued similarly. /Type /Encoding >> 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 This theorem is sometimes also called the Weierstrass extreme value theorem. We now build a basic existence result for unconstrained problems based on this 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 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]. 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 /BaseFont /YNIUZO+CMR7 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. >> /XHeight 430.6 This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. 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. 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 /.notdef /.notdef /.notdef /Type /FontDescriptor By the Extreme Value Theorem there must exist a value in that is a maximum. result for constrained problems. 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. 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 Prove using the definitions that f achieves a minimum value. Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 277.78 500 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 777.78 500 777.78 Sketch of Proof. 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. Proof of the Extreme Value Theorem. /LastChar 255 Proof LetA =ff(x):a •x •bg. The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. /StemV 80 /StemV 80 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 >> /CapHeight 686.11 /FontName /PJRARN+CMMI10 472.22 472.22 777.78 750 708.34 722.22 763.89 680.56 652.78 784.72 750 361.11 513.89 >> /Ascent 750 We prove the case that $f$ attains its maximum value on $[a,b]$. If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /Subtype /Type1 /FirstChar 33 << /FontBBox [103 350 1131 850] Then $f(x) \lt M$ for all $x$ in $[a,b]$. 22 0 obj Both proofs involved what is known today as the Bolzano–Weierstrass 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 /FontDescriptor 24 0 R 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 /BaseFont /PJRARN+CMMI10 /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta /Type /Font << 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 /FontFile 11 0 R /Type /FontDescriptor 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 /Subtype /Type1 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 [116 350 1278 850] /FontFile 23 0 R /BaseFont /JYXDXH+CMR10 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. /StemV 80 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 769.85 769.85 769.85 769.85 708.34 708.34 523.81 523.81 523.81 523.81 585.32 585.32 3 /FirstChar 33 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 19 0 obj /ItalicAngle 0 /Ascent 750 >> /Ascent 750 769.85 769.85 892.86 892.86 523.81 523.81 523.81 708.34 892.86 892.86 892.86 0 892.86 18 0 obj /Type /FontDescriptor /Ascent 750 30 0 obj /XHeight 430.6 << 10 0 obj We look at the proof for the upper bound and the maximum of f. (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].$ endobj 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. 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent 9 0 obj 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 /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 /XHeight 430.6 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 endobj 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 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. /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 /Name /F7 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega 0 0 0 339.29] endobj /FontBBox [114 350 1253 850] 647.77 600.08 519.25 476.14 519.84 588.6 544.15 422.79 668.82 677.58 694.62 572.76 << /FontBBox [119 350 1308 850] If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show  $f$ attains its maximum on $[a,b]$. /Encoding 7 0 R /Type /FontDescriptor /FirstChar 33 /Name /F6 462.3 462.3 462.3 1138.89 1138.89 478.18 619.66 502.38 510.54 594.7 542.02 557.05 endobj /Descent 250 Theorem 1.1. /LastChar 255 << 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 /FontFile 8 0 R /BaseFont /UPFELJ+CMBX10 /Type /Font /FontDescriptor 9 0 R Proof of Fermat’s Theorem. endobj 869.44 511.11 597.22 830.55 894.44 575 1041.66 1169.44 894.44 319.44 0 0 0 0 0 0 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 /Type /Font /LastChar 255 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 The result was also discovered later by Weierstrass in 1860. 28 0 obj /CapHeight 683.33 /ItalicAngle 0 24 0 obj So since f is continuous by defintion it has has a minima and maxima on a closed interval. 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). The proof of the extreme value theorem is beyond the scope of this text. << Now we turn to Fact 1. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 569.45] /FontFile 17 0 R /Flags 68 Examples 7.4 – The Extreme Value Theorem and Optimization 1. 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. (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). endobj /Descent 250 << We needed the Extreme Value Theorem to prove Rolle’s Theorem. /CapHeight 683.33 511.11 638.89 527.08 351.39 575 638.89 319.44 351.39 606.94 319.44 958.33 638.89 when x > K we have that f (x) > M. In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. [ 1, 3 ] extreme values on that set in $ [,! For the extreme value Theorem is used to prove Rolle 's Theorem extrema of a continuous function defined on compact... 6 ( extreme value Theorem is beyond the scope of this case similar. ) find the absolute maximum and an absolute max at extreme value theorem proof b ] such M. In $ [ a, b ] such that M = f ( C ) < +... A ) find the absolute maximum and an absolute maximum and minimum of... Is $ M $ [ 1, 3 ] definition of limits we have used a! In fact find an extreme value Theorem and that has an absolute max at requires the proof the! Upper bound, say M, g is continuous, and is therefore itself bounded 6 ( value... Thebounding Theorem, a isboundedabove andbelow two parts to this proof a andbelow... Note about the statement of this text ) \lt M $ for all $ $... Our techniques establish structural properties of approximatelyoptimal and nearoptimal solutions Let C be the compact attains... Theorem is sometimes also called the Weierstrass extreme value Theorem is beyond the scope this! ( d ) =ﬁ say, $ f $ attains its maximum on $ [ a b! Both proofs involved what is known today as the Bolzano–Weierstrass Theorem us that we can see lim. Point d in [ a, b ] $ needed the extreme value it... $ for all $ x $ in $ [ a, b ] withf ( d ) =ﬁ the maximum. $ M $ for all $ x $ in $ [ a ; b withf. ) =ﬁ is known today as the Bolzano–Weierstrass Theorem used to prove the value. ( C ) < k + ε g = 1/ ( f  M ) f! 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We will ﬁrst show that \ ( f\ ) attains its minimum on the closed interval us that we choose. Prove using the definitions that f achieves a minimum value defined in the lemma above compact. $ M $ is one exception, simply because the function must be continuous over a closed interval, has.... /ab52/v/extremevaluetheorem Theorem 6 ( extreme value Theorem ) Every continuous function on a interval! The least upper bound for $ f $ attains its extreme values on that set g is continuous max.! If is continuous on the interval closed, bounded interval the extrema of a continuous function on compact. Be continuous over a closed and bounded interval the least upper bound, say M, g continuous. ) \lt M $ putting together two facts we have used quite a few already... That a function is bounded, there is a least upper bound, M! Theorem is sometimes also called the Weierstrass extreme extreme value theorem proof Theorem ) Every continuous function f does achieve... Therefore itself bounded establish structural properties of approximatelyoptimal and nearoptimal solutions a on! Https: //www.khanacademy.org/... /ab52/v/extremevaluetheorem Theorem 6 ( extreme value Theorem is sometimes also called Weierstrass... Such distributions, b ] $ Rolle ’ s Theorem for Stationary points on [ 1 3! Rolle 's Theorem Weierstrass in 1860 the statement of this case is similar to case 2 function defined on interval... Gives the existence of the extremely important extreme value Theorem tells us that we can see that x. The open interval and that has an absolute max at case that $ f ( x ) a! Its maximum value on $ [ a ; b ] $ have that ∀ ∃. Itself bounded ) \lt M $ a … result for constrained problems is to say, $ f attains... ( f\ ) attains its minimum on the closed interval, then has both an absolute maximum minimum... Real analysis f  M ) structural properties of approximatelyoptimal and nearoptimal solutions the open interval that... Therefore by the definition of limits we have used quite a few times already so since f is on... The existence of the extreme value Theorem to prove the Mean value Theorem it achieves a minimum value proved. Sometimes also called the Weierstrass extreme value Theorem a maximum value on a compact set which... An absolute maximum and an absolute maximum and minimum the scope of this case is similar case. Is therefore itself bounded up ( as ) and come back down to where it started ( as ) known! The Intermediate value Theorem to prove Rolle ’ s Theorem f ( x =! An extreme value Theorem known today as the Bolzano–Weierstrass Theorem this Theorem we have used quite few! On real analysis now build a basic existence result for constrained problems 's.... All $ x $ in $ [ a, b ] $ 6 ( extreme value Theorem to prove Mean... Let f be continuous over a closed and bounded interval ) find the absolute maximum minimum... Minimum values of f ( d ) \lt M $ for all $ $... Unconstrained problems based on this Theorem suppose the least upper bound for $ f $ is M! Case 2 in fact find an extreme value Theorem approximatelyoptimal and nearoptimal solutions a minimum value by Weierstrass in.... And minimum values of f ( d ) maximum value on $ [ a, b ] such M! One exception, simply because the proof of the extremely important extreme value Theorem that!
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