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2311_IVT_EVT_MVT_figures&amp;solutions_F10

# 2311_IVT_EVT_MVT_figures&amp;solutions_F10 - MAC2311...

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Unformatted text preview: MAC2311' , Why Continuity and Differentiability Matter; Intermediate Value Theorem, Extreme value Theorem, and Mean Value Theorem Theorem: 'If f is differentiable at x = a, then f is continuous at x = a. Intermediate Value Theorem: Let f be continuous on the interval [ugh] , and let y be any number between f(a) and f (b) . Then there exists at least one number 0 between a and b such that f (c) a: y. Extreme Value Theorem: Let f be continuous on the interval [cab]. Then f has a global maximum and a global minimum on [61,13] . (Moreover, these global extrema must occur at _x = a or x = b or at a critical point between x: a and x=b .) .171 41135 ﬁgure, X 2:: C. :5 ﬂq e, (Mi? CHI/12a] 54an her in [63, E3] Mean Value Theorem: Suppose f is continuous on the closed interval [(1, b] and differentiable on the open interval (a,b). Then there exists at least one number c between a and I) such that , b — a f©=ﬂ)f0_ f b _ a F (a) l t. SloPe. 0'? Sierra, 0‘ hw‘jenil: Sf» moi‘ like a {in e a x:C Wroujln .R ) ((1 {RC0} and ([email protected] 1. Use the IVT to Show that p(x)=x3+x—l has aroot onthe interval [0,1]. r , k . . ,. . '5 State; (5:23)] WGM1QlS ewe LO¥1HHMOLLS or: GW)+OJ)) Pot)“: X +x~l ‘- ‘5' (ar‘l‘ctimlj Covi‘lﬁ‘muems m1 [0) l] we Since. Wit: 5 OCCLKV’ mkevéi :‘j 3‘ PVT-:0) we ‘2')“ jar—O (.1le observe "lMcCl' Pm): ml wlmle, PCI)“Z'-'—l ) 5:; 3'20 \5 + he'lweew {3(6) and PO) . 8:, the, inlermeclfal€VcLlue Wieovem ‘l‘lﬁre WW6 LG {1+ langl‘ one ﬂumlpar’ C he‘lfxﬁetaw Q MC) l jucL Hui" P(C) EEC-3 .- ‘Tk‘is Qwvctllwe is CL me)" “(we PHIL) . 2. Consider the function f(x)m(x—2)% =(3 x—2)2 shown, . , 2 _y 2 mth f(x)=g(x-2) 3=33 W' a) Circle all intervals below over which the hypothesis of the EVT is met. b) Circle all intervals below over which the hypothesis of the MVT is met. [1,10] 0) Consider f over [3,10] . Find the value of c guaranteed by the conclusion of the MVT. Draw the secant line through (3, f (3)) and (10, f (10)) , and then the tangent line at (c, f (0)). (They should look parallel!) a 43 2C) 5: --F(_IO) _ {3(3- te')=({f;?) 5! w , .y. 2 2: id... W3) =(é/mw2) “r seat: 7 I 2.. 3 ’2. WWW“... ,I He) e N52“: F335;“ ‘7 } nu...“- \ ‘4'" a “3’ 14) ~ . l ~A C5—— +2“ ”a“ 13..)3’ g highs?) (a . (3 Caz) ““ c1 Cwb'ﬂéa ...
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2311_IVT_EVT_MVT_figures&amp;solutions_F10 - MAC2311...

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