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. /Flags 68 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 9 0 obj 15 0 obj /Type /Font /FontBBox [-134 -1122 1477 920] Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. Since both of these one-sided limits are equal, they must also both equal zero. /ItalicAngle 0 Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. /Type /Font /Name /F2 /FontName /IXTMEL+CMMI7 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. 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 >> /Type /FontDescriptor 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. Proof LetA =ff(x):a •x •bg. 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 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 0 0 0 0 0 0 277.78] /FontFile 14 0 R /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring /ItalicAngle -14 /CapHeight 686.11 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /Type /Font butions requires the proof of novel extreme value theorems for such distributions. This theorem is sometimes also called the Weierstrass extreme value theorem. endobj 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). 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 /Flags 4 endobj /BaseFont /PJRARN+CMMI10 /FontDescriptor 21 0 R That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. 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. 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 777.78 625 916.67 750 777.78 680.56 777.78 736.11 555.56 722.22 750 750 1027.78 750 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 Proof of the 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 7 0 obj Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. 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 (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. /Name /F1 646.83 970.24 970.24 323.41 354.17 569.45 569.45 569.45 569.45 569.45 843.26 507.94 Therefore by the definition of limits we have that ∀ M ∃ K s.t. /CapHeight 686.11 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 /FirstChar 33 /Subtype /Type1 /Length 3528 /Subtype /Type1 Therefore proving Fermat’s Theorem for Stationary Points. /FontBBox [-100 -350 1100 850] /XHeight 430.6 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 We look at the proof for the upper bound and the maximum of f. 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 625 500 625 513.31 343.75 562.5 625 312.5 343.75 593.75 312.5 937.5 625 562.5 625 Suppose the least upper bound for $f$ is $M$. /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. /Flags 4 /FontFile 23 0 R endobj /FontFile 11 0 R /Ascent 750 Sketch of Proof. /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta /FontBBox [-116 -350 1278 850] /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 /.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 0 562.5] /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 /FirstChar 33 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent 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. /FontBBox [-103 -350 1131 850] State where those values occur. /StemV 80 28 0 obj 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 /StemV 80 f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. /FontName /PJRARN+CMMI10 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 /LastChar 255 /BaseEncoding /WinAnsiEncoding /FontDescriptor 15 0 R 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 0 0 0 339.29] So since f is continuous by defintion it has has a minima and maxima on a closed interval. /LastChar 255 << 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 0 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 /FontDescriptor 27 0 R /Subtype /Type1 /FirstChar 33 We needed the Extreme Value Theorem to prove Rolle’s Theorem. /StemV 80 24 0 obj /XHeight 430.6 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. 1138.89 339.29 339.29 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 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 /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] 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 $[a,b]$. 472.22 472.22 777.78 750 708.34 722.22 763.89 680.56 652.78 784.72 750 361.11 513.89 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 25 0 obj /XHeight 430.6 /Descent -250 Sketch of Proof. As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. when x > K we have that f (x) > M. /BaseFont /UPFELJ+CMBX10 /Encoding 7 0 R 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. /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright /CapHeight 683.33 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. https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem We now build a basic existence result for unconstrained problems based on this theorem. 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. endobj /XHeight 444.4 0 0 0 0 0 0 575] Among all ellipses enclosing a fixed area there is one with a … /Descent -250 k – ε < f (c) < k + ε. /BaseFont /TFBPDM+CMSY7 /Subtype /Type1 /LastChar 255 Since the function is bounded, there is a least upper bound, say M, for the range of the function. Indeed, complex analysis is the natural arena for such a theorem to be proven. Now we turn to Fact 1. We prove the case that $f$ attains its maximum value on $[a,b]$. << 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 Weclaim that thereisd2[a;b]withf(d)=fi. Typically, it is proved in a course on real analysis. 569.45] (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. 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 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /FontName /UPFELJ+CMBX10 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]. 10 0 obj /ItalicAngle -14 /FirstChar 33 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 ]. 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. 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 /Type /Font endobj Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. 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 /LastChar 255 The result was also discovered later by Weierstrass in 1860. /XHeight 430.6 Theorem 6 (Extreme Value Theorem) Suppose a < b. 21 0 obj /BaseFont /YNIUZO+CMR7 /FirstChar 33 >> 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 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 Prove using the definitions that f achieves a minimum value. ∀ M ∃ k s.t C be the compact set on which seek... 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