endobj 0 892.86] Prove using the definitions that f achieves a minimum value. 3 /CapHeight 683.33 Proof: There will be two parts to this proof. First we will show that there must be a ﬁnite maximum value for f (this State where those values occur. /Encoding 7 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 Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. Consider the function g = 1/ (f - M). 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. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /FontFile 20 0 R /Descent -250 /Type /FontDescriptor /Flags 4 /FirstChar 33 /FontName /PJRARN+CMMI10 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. /BaseFont /TFBPDM+CMSY7 24 0 obj Also we can see that lim x → ± ∞ f (x) = ∞. /Descent -250 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. 511.11 638.89 527.08 351.39 575 638.89 319.44 351.39 606.94 319.44 958.33 638.89 << endobj endobj 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. The proof of the extreme value theorem is beyond the scope of this text. << Therefore proving Fermat’s Theorem for Stationary Points. /FontDescriptor 27 0 R The proof that \$f\$ attains its minimum on the same interval is argued similarly. /BaseFont /UPFELJ+CMBX10 Since f never attains the value M, g is continuous, and is therefore itself bounded. << /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave /acute /caron /breve /macron /ring << /Flags 4 (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. endobj If a function \$f\$ is continuous on \$[a,b]\$, then it attains its maximum and minimum values on \$[a,b]\$. /StemV 80 0 0 0 339.29] 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. Since both of these one-sided limits are equal, they must also both equal zero. /FirstChar 33 /Type /FontDescriptor Therefore by the definition of limits we have that ∀ M ∃ K s.t. /FontName /UPFELJ+CMBX10 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. That is to say, \$f\$ attains its maximum on \$[a,b]\$. 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 We now build a basic existence result for unconstrained problems based on this theorem. 30 0 obj /Type /Font 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 0 0 0 0 0 0 277.78] /FontDescriptor 15 0 R /Widths [342.59 581.02 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 /FontName /YNIUZO+CMR7 /BaseEncoding /WinAnsiEncoding >> 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. We prove the case that \$f\$ attains its maximum value on \$[a,b]\$. /Name /F1 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 << Then the image D as defined in the lemma above is compact. /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 828.47 580.56 682.64 388.89 388.89 388.89 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 The Mean Value Theorem for Integrals. /Subtype /Type1 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 /FontBBox [-103 -350 1131 850] Letﬁ =supA. /FontDescriptor 18 0 R /XHeight 430.6 594.44 901.38 691.66 1091.66 900 863.88 786.11 863.88 862.5 638.89 800 884.72 869.44 , and is therefore itself bounded to say, \$ f \$ attains its maximum on [! Must go up ( as ) ) find the absolute maximum and minimum values f. 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